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The thermodynamics of hysteresis for high pressure metal hydrides
Journal of the Less-Common
Metals, 125 (1986) 241 - 260
THE THE~ODYN~ICS METAL HYDRIDES TED B. FLANAGAN Chemistry Department, CHOONG-NYEON Department
247
OF HYSTERESIS FOR HIGH PR~SS~RE
and II. S. CHUNG ~niue~i~
of Vermont. 3urlingt~n,
VT 05405
(U.S.A.)
PARK
of Metallurgy, Chonnam National University, Kwangiu, Chonnam
(Korea)
(Received April 8, 1986; in revised form July 7,1986)
Summary The thermodynamics of hysteresis for metal hydrides is developed for the regime of high hydrogen pressures where the effect of hydrostatic pressure on the solid phases and deviations from ideal gas behavior must be considered. In contrast with the low pressure behavior, at high pressures the calorimetric enthalpies for hydride formation and decomposition are expected to differ significantly even when the phase boundary compositions are not greatly affected by hysteresis. An expression for the entropy production due to hysteresis is derived for the high pressure regime. A general explanation is offered for the large hysteresis which usually accompanies hydride formation and decomposition at high pressures.
1. Introduction Most metal hydrides exhibit hysteresis. This means that under isothermal conditions the pressure needed to form the hydride phase is greater than the pressure for the decomposition of the hydride phase. This leads to losses of efficiency in the application of metal hydrides. Some metal hydrides can be formed only at pressures of hydrogen greatly exceeding 1 atm, e.g., NiH, RhH etc. The preparation and ch~ac~rization of such high pressure hydrides has been pioneered by Baranowski and his coworkers [ 13. On the basis of results for many high pressure hydrides it has been noted that hysteresis for high pressure hydrides is especially large [ 21. Recently Tkacz and Baranowski [2] have reported the following calorimetric results at high pressures for the formation and decomposition of NiH: AH,(cal) = -5.02 kJ (mol Hf-’ at Pf = 6200 bar and AH&Cal) = 3.13 kJ (mol H)-’ at pd = 3400 bar at 298 K. They corrected these values to conditions of 1 atm giving AHT(cal) = -8.14 kJ (mol H)-l and Lwli(cal) = 4.48 kJ (mol H)-‘. For this correction they employed 0022-5088/86/$3.50
0 Elsevier Sequoia~Printed in The Netherlands
248
where values of on, and VH, are available over the required pressure ranges [3, 41. They did not include the effect of pressure on the solid phases because it was considered to be small in comparison with the non-ideality of the gaseous hydrogen. It has been pointed out that provided the forward and reverse reactions are nearly the same, i.e. the phase boundary compositions are the same during hydride formation and decomposition, Lw values determined calorimetrically for the two-phase region should be unaffected by hysteresis [ 51. However, LW values derived from van’t Hoff plots of the plateau pressure for hydride formation pf or for hydride decomposition pd will be affected by hysteresis and ]&&,I > lmf]. Unfortunately there are not many precise calorimetric data available to confirm this prediction but there seems to be no reason to doubt its validity for the low pressure regime because it is firmly based on thermodynamics [5]. It should be pointed out that this thermodynamic development was restricted to the low pressure regime where hydrogen behaves ideally and the effect of hydrostatic pressure on the solid phases is nil [ 51. It is clear from the results of Tkacz and Baranowski [2] that this prediction may not be valid for the high pressure regime. The purpose of the present work is to extend the thermodynamic treatment of hysteresis given earlier to the high pressure regime and then to compare its predictions with the results of Tkacz and Baranowski. In addition we will address the general question of why hysteresis is found to be large for high pressure hydrides [ 21. It should be emphasized that hysteresis is particularly important for high pressure hydrides because it is a large fraction of the measured two-phase enthalpies since values of IMI are rather small for high pressure hydrides and hysteresis is relatively large.
2. Thermodynamic hydrides
treatment of isothermal hysteresis for high pressure
Figure 1 shows a schematic hysteresis cycle for a system consisting of one mole of hydrogen distributed between one mole of metal and the gaseous phase. The system is enclosed in a constant temperature reservoir with a frictionless piston enabling pressure changes to be carried out slowly and reversibly, at least for those steps which do not involve any inherent irreversibility. For simplicity it has been assumed that the hydride composition is [H] / [M] = 1 and the dilute phase composition is [H] /[Ml = 0, i.e. the reaction is M(s) + $H,(g) = MH(s). This is not a bad approximation for the Ni-H system at 298 K [ 21. pes in Fig. 1 is the hypothetical equilibrium pressure which is believed to lie about half way between pf and pd [6] but, in any case, iS somewhere between pf and pd. For each pressure there is a corresponding fugacity which is assumed to be known [ 3, 41.
249
5
4 1
H/N-
C
Fig. 1. Schematic
representation
of an isothermal
hysteresis
cycle.
The free energy changes for the irreversible steps, i.e. those which give rise to hysteresis regardless of how slowly the changes are carried out, 1 + 2 and 4 + 5 can be evaluated by following reversible pathways [7]. For example, for step 1 -+ 2 AG(l
-+ 2) =
AG(l -+2) =
AG(l + 6) + AG(6 + 3) + AG(3 --f 2) pw s
(1)
v
3
Pf
2
dp + AG,,(=
0) + jf
Vn dp
(2)
peq
where V, and V, are the partial molar volumes and VH, is the molar volume of hydrogen gas. From experimental results on metal-H systems it is a reasonable approximation to consider V, and V, independent of hydrogen content. V, therefore cancels between steps 1 + 6 and 3 -+ 2 and does not appear in eqn. (2). By following similar reversible pathways the step corresponding to hydride decomposition is: AG(4 + 5) = id
2
dp + 4’” Vn dp
Peq
(3)
pd
Steps 1 + 2 and 4 + 5 are the only irreversible steps in the cycle so that the free energy dissipated in the cycle is the sum of AG( 1 -+2) and AG(4 --f 5) given by eqn. (4) -J
diG = Jf
J$
dp - Jf VH dp = RT ln[$l”2
pd
- V&f
-pd)
(4)
Pd
The enthalpy and entropy changes can be evaluated in the same manner by following reversible paths. We obtain M(l+
2) =
Peq V 5 s Pf
(1- oH2T) dp - lAI&,I + J'Vn(1 - or,T) dp Peq
(5)
250 Azq4 9 5) = idJ$l Peq
-%$f’)
dp + lMH,,I + /peqVH(l -aHT)
dp
(6)
J'd
where on = VH-1(t3VH/XZ’),. The sum of AH(1 + 2) and AH(4 + 5) is m(l+2)+m(4+5)=--f
~(l-aH,T)dp+yfv~(l--oIHndp pd
(7)
pd
For the entropy changes we obtain p-2
AS(l-+2)=-
~~~,d~-]~pq]-~~oi~~~d~
s
Pf
jd+
Mt4 + 5) = -
(8) peq
CY~,
dp + ]A$,]
- i’”
aHVH dp
(9)
Pd
Peq and ds(1 + 2) + AS(4 + 5) = if 2
CYn,dp - ~~~Hv~ dp
pd
fW
J’d
Generally the entropy change can be written as dS = diS + d,S
(11)
where diS is the entropy produced in the system and d,S is the entropy change in the system as reflected by heat transfer with the surroundings. Forstep1-+2 (12)
AiS(1+ 2) = AS(l + 2) - A,S(l + 2)
where A$(1 -+ 2) = Q(1 + 2)/T = AH( 1 + 2)/T and therefore using eqns. (5) and (8), AiS(l -+ 2) can be written as
A,s(l+2)=j* 52 c&-f*: Peq
2T
(13)
dp
P-J
and similarly AiS(4 + 5) can be obtained from eqns. (6), (9) and (11) Pea
AiS(4 + 5) = $ Pd
Per2
2
dp-$
2
dp
(14)
Pd
and since for the reversible steps 2 + 4 and 5 + 1 AiS is zero, we obtain for the cycle
251
jd,S=[!!$
a-i’:
dp=--_!_J
d,G=iRln
i
-
vH(pFpd)
pd
(15) Although $d$ and $dfG are not zero for the cycle, $ ds, $ dG and $ d.H must, of course, be zero for the complete cycle. This means that _#d,S exactly cancels f diS and the changes AH( 1+ 2) + AH(4 + 5) are exactly canceled by the enthalpy changes which occur during the two reversible steps 2 + 4 and 5 -+ 1. It is of interest to calculate the reversible work done on the system by the surroundings during the complete cycle. $dU’=jf2
(16)
dp-_jfVHdp pd
pd
The work done on the solid cancels in steps 5 + 1 and 2 -+ 4 but not in steps 1 + 2 and 4 * 5. As for the low pressure regime, the net work done on the system is equal to the free energy dissipation in the cycle (eqn. (4)) and the net heat dissipated to the surroundings, i.e. JdQ=-JdW=-J’%
dp+ pd
3. Thermodynamic measurements
JfVndp=JdiG
(17)
Pd
parameters
region derived from p-c-T
for the two-phase
In the low pressure regime thermodynamic parameters can be derived from the temperature dependence of pf and pd but in the high pressure regime these plateau pressures must be converted to fugacities in order to derive thermodynamic parameters. is not known with certainty, it is, none the less, Although P&,) informative to consider the value of the enthalpy change which would be derived from its temperature dependence. Next we will consider the analogous enthalpies derived from pf(ff) and p&d) and their relationships to the enthalpy change derived from p,,(f,,). For hydride formation at peq AG(5 -f 6) = AG,Cp,,)
-
pw
s
= 0 = AH;-
TAS;+
s
vdl-
~HT)
dp
peq
v
T(i-cru,T)dp+TJ
1
oHVHdp 1
peq v -TJ
+Ynldp 1
(18)
253
From eqn. (2), l/2
- V&,
- pes) = -AG(
1 --f 2) = wf
where wp is the plastic, irreversible work done on the solid phase needed to accomodate the volume changes during hydride formation [ 5, 61. The analogous quantity for hydride decomposition is wpd.These irreversible work terms are assumed to be independent of temperature at least over the temperature range where the phase boundary compositions do not change significantly with temperature. From eqn. (25), R 3 In
i
ff”’
aT-’
i
apf
= LWi(app) = -lAWI
+ wg + VHpf + V, T aT-’
This can be expressed in terms of 1kI&(app)l
~Sw)
= -I ~&(wp)l
+4 +
vH
r
using eqn. (22)
ati, -peq) aT-'
+ VH(Pf
Proceeding in a similar way for hydride decomposition, obtain _-R
(26)
-Peq)
(27)
step 4 + 5 we
a In
fd1j2 v, ap, = AKi(wp) = lAIJH”I- V,p, - r aT-’ + wpd i aT-’ I
or expressing this in terms of I AK&(app)l
A-Ki(app)= I~~,(app)I
+ wX +
VH
r
, (eqn. (22)),
abeq-Pd) aT-'
gives
+ VHb-bq
-Pd)
(29)
It should be emphasized that hysteresis affects the derived values of Mi(app) and U”,(app) significantly because at high pressures hysteresis is a large fraction of the AH’ values. It should also be noticed that the determination of M” from AW(app) by eqns. (26) and (29) is not straightforward because the correction terms are comparable with the enthalpy values.
4. High pressure hysteresis for nickel hydride The values of the calorimetric enthalpies determined by Tkacz and Baranowski [2] for nickel hydride have been noted in the introduction. Now the calorimetric enthalpies for hydride formation and decomposition correspond to eqns. (5) and (6) because the heats detected by the calorimeter are equal to the enthalpy changes under constant pressure conditions. Therefore for Ni-H
254
p-l
A&(d)
= AH(1 -+ 2) =
v
J
+ j’V,dp
+(l-on,T)dp-lAZY~ql
JJf
-5.02
=
A&(cal)
= M(4 =
Peq
kJ (mol H)-’
+ 5) = j”? PW
(30)
(1 -cuulT)
dp + ]Lw,,I
+ j”
Vu dp
pd
3.13 kJ (mol H)-’
(31)
where the uncertainty in these quantities is reported to be +O.l kJ (mol H)-r . From these two values we obtain for the sum: AEZ(1 -+2) + M(4 + 5) = -1.89 + 0.2 kJ (mol H)-‘. From eqn. (7) and the known properties of H,(g) [ 3, 41 we obtain
m(l+2)+m(4+5)=-j$l-C$T)dp+@,--p,)Vu Pd
= -1.96
+ 0.48 = -1.48
kJ (mol H)-’
where it has been assumed that VH = 1.7 X lop6m3 (mol H)-’ [8] and independent of pressure. The agreement with experiment is not bad considering that VH may be somewhat smaller at [H]/[Ni] - 1 than 1.7 X 10e6 m3 (mol H)-’ [ 81. Next the free energy dissipation due to hysteresis will be evaluated for a real high pressure hydride - the Ni-H system. From eqn. (4)
-J
2
diG = if
dp - VH(Pf -Pd)
= RT ln(+)“2
- VH(JIf -Pd)
Pd
= 2.71 - 0.48 = 2.23 kJ (mol H)-’ where it has again been assumed that VH = 1.7 X 10e6 m3 (mol H)-’ independent of pressure. The values used for evaluation of $ diG are taken from experimental results [ 23. This hysteresis is greater than twice that observed for Pd-H in the low pressure regime. This value for Ni-H can be compared with a value calculated from eqns. (7) and (lo), i.e. - f diG =-{IvI(l
+ 2) + M(4
+ 5)) - T(AS(1 + 2) + AS(4 + 5))
Pf
=-
s
?(l-
Pd
Pf
%,T)dP+@f-Pd)VH-
V J -+-++P Pd
(32)
255
The term onVn dp has been dropped from aS(l + 2) + AS(4 + 5) because it is small. For the enthalpy terms we take -1.89 + 0.2 kJ (mol H)-’ which is the experimental value from the calorimetric results [2] and the last term can be evaluated from the known properties of H,(g). We obtain -$ diG = +2.62 kJ (mol H)-’ which is somewhat greater than that found from experimental values of pf and pd, but it is probably within experimental error. So far only thermodynamic parameters for the hysteresis cycle have been considered and these are independent of assumptions about the location of equilibrium. Thermodynamic parameters for hydride formation and decomposition will now be evaluated and for this purpose an assumption about the location of equilibrium is needed. We assume that f,,* * fffd which leads to the following for Ni-H (2): pes = 4675 bar (f,, = 8.5 X 104) using the experimental values pi = 6200 bar (ff = 2.43 X 105), pd = 3400 bar (fd = 2.97 X 104). Using these values and eqns. (5) and (6), we obtain Mf(cal)
= -5020
PeqVH
J (mol H)-’ =
s +
from which we obtain IA&,( sition Md(d)
= 3130 J
(mOl
aHIT)
2(l-
Pf (Pf
dp
-
b-&,I
(33)
--Pe.dVH
= 4150 J (mol H)-l and for hydride decompo-
H)-l = j”
‘2
(1 -on,T)
dp + IA&,1
Peq +
be,
(34)
-PdlVH
from which we obtain IMe, = 3970 J (mol H)-‘. The values of 1AI&.,1 should be nearly identical and they are, considering the experimental error of the calorimetric values [2]. If the calorimetric values are as accurate as stated, this good agreement lends support for the assumption that f,,* = fffd. The plastic deformation which occurs for both hydride formation and decomposition is therefore of about equal magnitude so that wp = wpd. The entropy changes M(1 -+ 2) at pf and AS(4 + 5) at pd can be calculated for Ni-H from eqns. (1) and (5) for formation (1 -+ 2) and eqns. (2) and (6) for decomposition (4 + 5). For example, at T = 298 K
As(l+ 2) =
U(l -+2) T
+ $Pf-Peq)
-
AG(lT--, 2) = m$cal)
=-
5020
T
_
+1115= -13.1 T
2
dp
J K-’ (mol H))’
and for aS(4 + 5) = 14.24 J K-’ ( mol H))‘. These magnitudes should differ by
256 pi
s
2
on, dp = 2.5 J K-l (mol H)-’
J’d
and they differ by somewhat less than this but their difference is within experimental error.
5. Correction of thermodynamic parameters to 1 atm Tkacz and Baranowski [ 21 corrected their high pressure enthalpy values to values expected at 1 atm. They employed the equation given in the Introduction to correct formation and decomposition values. It can be seen from eqns. (5) and (6) that effects of hysteresis should disappear in the calorimetric enthalpies, m(l --f 2) and M(4 + 5), as p + 1 atm, if, as assumed, the forward and reverse reactions are identical. We correct the observed values as follows:
= -5020
f 4409 - 1054 = -1665
J (mol H)-l (35)
pd c aH,T)
dp
+ v,Pd
= 3130 - 2460 + 578 = 1248 J (mol El)-’ Thus it can be seen that the magnitudes of the two enthalpies are reasonably close. This supports the present analysis because we have argued that the influence of hysteresis on the enthalpies determined calorimetrically should go to zero as p -+ 1 atm. Tkacz and Baranowski [2] give quite different values for 1 atm conditions than those derived here; they omitted the hydrostatic terms for their correction to 1 atm but there also appears to be an error in the sign of the correction term with respect to the sign of their A& and A& values both of which are reported as positive. It is interesting that the high pressure conditions lead to a more exothermic enthalpy of hydride formation as compared with that at 1 atm; this is one of the factors that allows the hydride phase to form despite the positive mH0, values (see below). It is also of interest to calculate the entropy change expected at 1 atm
= -13.6
- 35.1 = -48.7
J K-’ (mol H)-”
257
This 1 atm calculated value is quite reasonable because, e.g., for Pd-H, AS”,= -46 J K-’ (mol H)-i and it has been noted that there is not much variation amongst aS(; values for various metals. For Ni-H in passing from conditions of 1 atm to peq, AH decreases by 2.6 kJ (mol H)-’ and --T AS decreases by 10.5 kJ (mol H))’ and thus both terms contribute to the formation of the hydride phase at high pressures. As(l+ 2) and A@ 4 + 5) can also be corrected to conditions of 1 atmH,. We have aSi = AS{1 + 2) - If
2
ou, dp = -49.4
J K-’ (mol H)-’
1 &i$
= As(4 -+ 5) -j
?
on, dp = 48.03 J K-i (mol H)-l
pd
These two values differ slightly but their average is close to I AS& I = 48.7 J K-l (mol H))‘. This demonstrates that hysteresis does not affect the standard entropy values much. It should be emphasized, however, that ASi or AS’: (1 atm) should not be calculated from AH( 13 Z)(cal)/!I’ - R In ff”2 or m(4 + 5)(cal)/T + R In fd1’2. These equations give -68.8 J K-’ (mol H))’ and +53.3 J K-’ (mol H)-’ respectively for Ni-H. This is not correct as we have shown before for the low pressure regime 151. Antonov et al. [9] measured pf and pa values for Ni-H over a wide temperature range and these data seem to be compatible with the 298 K data of Tkacz and Baranowski [ 21. In order to be able to employ these data to derive values of ~~(app) from the ~mperature dependence of ff and fd (eqns. (26) and (28)) reliable relationships between f and p must be available for hydrogen above 307 K, which is the limit of the results of Mills et al. [3]. Fukai and Sugimoto [lo] have corrected the results of Mills et al. f33 in order for them to apply at temperatures above 307 K. Using Table 1 in the paper of Fukai and Sugimoto [lo] we have derived from the data of refs. 2 and 9, U”f(app) = -3.1 kJ (mol H)-l and mi(app) = 5.9 kJ (mol H))’ from plots of In ff1’2 and In fd1’2against T-l. From eqns. (26) and (28) and using the values of A@ derived in calculations (35) we obtain (in the order of terms which appear in eqns. (26) and (28)): A.E?i(app) = -116651 oJii(app)
+ 1530 - 1960 + (223012) = -1454
= + I12481 - 1020 + 1884 + (2230/2)
J (mol H)-’
= 3227 J (mol H)-’
The agreement of the calculated values with those determined from the experimental high pressure data is not very good but this may be due largely to the errors in the pressure-fugacity relationships above 307 K. In any case it is seen that in both cases the absolute value of the decomposition enthalpy is greater than the value for hydride formation. In this section we wish to make the point that enthalpies (and entropies) derived from van% Hoff plots for high pressure hydrides do not give the 1 atm values nor the high pressure
258
calorimetric values, and to obtain the 1 atm values from the apparent values requires correction terms which are comparable with the values themselves. 6. Why is hysteresis large for high pressure hydrides? It has been noted by Baranowski and coworkers in their research on high pressure hydrides that hysteresis is invariably large for these systems as compared with low pressure hydrides. No general explanation of this has been given although Baranowski and Bochenska [ll] have discussed the specific case of Ni-H using the model proposed by Scholtus and Hall [12] for hysteresis. It should be pointed out that it has not been possible to predict quantitatively the hysteresis expected for specific low pressure systems so that to expect a quantitative explanation for the high pressure hydrides is unrealistic, however, we will attempt to indicate in a general way why high pressure hydrides have large hysteresis. It should first be pointed out that hysteresis is not given by RT In(ff/ fd)1’2 for high pressure hydrides but the term V&Q -pd) must be subtracted from it (eqn. (4)) to give -$ diG. Often hysteresis in these high pressure hydrides is assumed to be given only by the RT l~~(f~/f~)“~ term but even after V&I, -pd) is subtracted from RT ln(ff/fd)1’2, the values of -9 diG are still quite large for high pressure hydrides and an explanation is provided in the following. Welch and Pick [13] have given a correlation between MU, and @/a where A& is the heat of solution of hydrogen at infinite dilution of hydrogen, p is the shear modulus and Q is the atomic volume of the metal The correlation shows thai MH”, increases with increasing p/a. Amongst these metals the largest variation comes from p rather than C?. It is clear from an examination of the literature that AE?& values for high pressure hydrides are greater than for low pressure hydrides. For example, for Ni-H, A.?&= 27 kJ (mol H)-i [14] and for Pd-H, AZY$ = -10.2 kJ (mol H)-i [15]. According to the regular interstitial solution model [ 161 RT In peq1’2 = AG: = AH& - T AS: + h1/2 where h, is the H-H interaction parameter which is negative and comparable for most systems. This simple model, although not quantitatively satisfactory, suffices to show how the standard free energy for hydride formation depends on A.?& it should be noted that AS; is similar for most systems. Now if LM”, is large, it follows from the correlation between AiYk and or [13] that p is large. Hysteresis in metal hydrides is believed to be due to the irreversible plastic work needed to accomodate the volume changes [6] and the energy needed for this plastic work is expected to be proportional to the energy for dislocation formation which, in turns, is proportional to ,V [ 171. Hence it follows that hysteresis should be large for high pressure hydrides. Let us apply these considerations quantitatively to Ni-H. For Ni-H, -$ diG = 2.23 kJ (mol H)-i [2]. For Pd-H, -9 diG = 0.850 kJ (mol H)-’
259
[6]. The abrupt lattice change is smaller for Pd-H than for Ni-H and therefore hysteresis should be larger for the latter by the factor: (b - U)Ni-n/ (b - c_I)~,_-~= 1.66. The shear modulus of nickel is greater than that for palladium by a factor of 1.47 [18]. These two factors should influence hysteresis multiplicatively and therefore -$ diG should be (1.66) (1.47) (0.850) kJ (mol H)-’ = 2.08 kJ (mol H)-’ in comparison with the experimental value of 2.33 kJ (mol H)-l which is reasonably good agreement. One further example will be considered and this is the PdRh (XRh = 0.80) alloy where the hysteresis is quite large [ 191. For this alloy -$ diG = 4.50 kJ (mol H)-l. Now proceeding in the same way as for Ni-H we have: (1.66) (2.5) (0.85) kJ (mol H)-l = 3.52 kJ (mol H)-’ where the shear modulus has been taken to be equal to the moduli of the pure elements times the mole fraction of each element in the alloy. The agreement is not as good as for the case of Ni-H but the plateau pressure for hydride formation is not as well defined as for the case of Ni-H but a large hysteresis is predicted. We have avoided considering any high pressure hydrides where there is a change of structure on formation of the hydride phase, e.g. Mn-H [20], because hysteresis may be more complex in these cases. For the two hydrides considered above there is merely an abrupt lattice expansion or contraction on hydride formation or decomposition.
7. Conclusions In the present paper we have shown that the entropy production for a high pressure hysteresis cycle is given by eqn. (15). In contrast with conditions of normal pressures, where ideal gas behavior can be assumed and hydrostatic effects on the solid phases are insignificant, the enthalpies for hydride formation and decomposition measured calorimetrically differ significantly at high pressures. The derived equations for high pressure hysteresis are employed for the Ni-H system. An explanation is offered for the fact that hysteresis appears to be large for high pressure hydrides as compared with normal hydrides which form and decompose at pressures below about 100 atm. Acknowledgments TBF and HSC wish to acknowledge National Science Foundation.
the financial
support
of the
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in G. Alefeld
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Hydrogen
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Springer,
Berlin,
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Metals, 125 (1986) 241 - 260
THE THE~ODYN~ICS METAL HYDRIDES TED B. FLANAGAN Chemistry Department, CHOONG-NYEON Department
247
OF HYSTERESIS FOR HIGH PR~SS~RE
and II. S. CHUNG ~niue~i~
of Vermont. 3urlingt~n,
VT 05405
(U.S.A.)
PARK
of Metallurgy, Chonnam National University, Kwangiu, Chonnam
(Korea)
(Received April 8, 1986; in revised form July 7,1986)
Summary The thermodynamics of hysteresis for metal hydrides is developed for the regime of high hydrogen pressures where the effect of hydrostatic pressure on the solid phases and deviations from ideal gas behavior must be considered. In contrast with the low pressure behavior, at high pressures the calorimetric enthalpies for hydride formation and decomposition are expected to differ significantly even when the phase boundary compositions are not greatly affected by hysteresis. An expression for the entropy production due to hysteresis is derived for the high pressure regime. A general explanation is offered for the large hysteresis which usually accompanies hydride formation and decomposition at high pressures.
1. Introduction Most metal hydrides exhibit hysteresis. This means that under isothermal conditions the pressure needed to form the hydride phase is greater than the pressure for the decomposition of the hydride phase. This leads to losses of efficiency in the application of metal hydrides. Some metal hydrides can be formed only at pressures of hydrogen greatly exceeding 1 atm, e.g., NiH, RhH etc. The preparation and ch~ac~rization of such high pressure hydrides has been pioneered by Baranowski and his coworkers [ 13. On the basis of results for many high pressure hydrides it has been noted that hysteresis for high pressure hydrides is especially large [ 21. Recently Tkacz and Baranowski [2] have reported the following calorimetric results at high pressures for the formation and decomposition of NiH: AH,(cal) = -5.02 kJ (mol Hf-’ at Pf = 6200 bar and AH&Cal) = 3.13 kJ (mol H)-’ at pd = 3400 bar at 298 K. They corrected these values to conditions of 1 atm giving AHT(cal) = -8.14 kJ (mol H)-l and Lwli(cal) = 4.48 kJ (mol H)-‘. For this correction they employed 0022-5088/86/$3.50
0 Elsevier Sequoia~Printed in The Netherlands
248
where values of on, and VH, are available over the required pressure ranges [3, 41. They did not include the effect of pressure on the solid phases because it was considered to be small in comparison with the non-ideality of the gaseous hydrogen. It has been pointed out that provided the forward and reverse reactions are nearly the same, i.e. the phase boundary compositions are the same during hydride formation and decomposition, Lw values determined calorimetrically for the two-phase region should be unaffected by hysteresis [ 51. However, LW values derived from van’t Hoff plots of the plateau pressure for hydride formation pf or for hydride decomposition pd will be affected by hysteresis and ]&&,I > lmf]. Unfortunately there are not many precise calorimetric data available to confirm this prediction but there seems to be no reason to doubt its validity for the low pressure regime because it is firmly based on thermodynamics [5]. It should be pointed out that this thermodynamic development was restricted to the low pressure regime where hydrogen behaves ideally and the effect of hydrostatic pressure on the solid phases is nil [ 51. It is clear from the results of Tkacz and Baranowski [2] that this prediction may not be valid for the high pressure regime. The purpose of the present work is to extend the thermodynamic treatment of hysteresis given earlier to the high pressure regime and then to compare its predictions with the results of Tkacz and Baranowski. In addition we will address the general question of why hysteresis is found to be large for high pressure hydrides [ 21. It should be emphasized that hysteresis is particularly important for high pressure hydrides because it is a large fraction of the measured two-phase enthalpies since values of IMI are rather small for high pressure hydrides and hysteresis is relatively large.
2. Thermodynamic hydrides
treatment of isothermal hysteresis for high pressure
Figure 1 shows a schematic hysteresis cycle for a system consisting of one mole of hydrogen distributed between one mole of metal and the gaseous phase. The system is enclosed in a constant temperature reservoir with a frictionless piston enabling pressure changes to be carried out slowly and reversibly, at least for those steps which do not involve any inherent irreversibility. For simplicity it has been assumed that the hydride composition is [H] / [M] = 1 and the dilute phase composition is [H] /[Ml = 0, i.e. the reaction is M(s) + $H,(g) = MH(s). This is not a bad approximation for the Ni-H system at 298 K [ 21. pes in Fig. 1 is the hypothetical equilibrium pressure which is believed to lie about half way between pf and pd [6] but, in any case, iS somewhere between pf and pd. For each pressure there is a corresponding fugacity which is assumed to be known [ 3, 41.
249
5
4 1
H/N-
C
Fig. 1. Schematic
representation
of an isothermal
hysteresis
cycle.
The free energy changes for the irreversible steps, i.e. those which give rise to hysteresis regardless of how slowly the changes are carried out, 1 + 2 and 4 + 5 can be evaluated by following reversible pathways [7]. For example, for step 1 -+ 2 AG(l
-+ 2) =
AG(l -+2) =
AG(l + 6) + AG(6 + 3) + AG(3 --f 2) pw s
(1)
v
3
Pf
2
dp + AG,,(=
0) + jf
Vn dp
(2)
peq
where V, and V, are the partial molar volumes and VH, is the molar volume of hydrogen gas. From experimental results on metal-H systems it is a reasonable approximation to consider V, and V, independent of hydrogen content. V, therefore cancels between steps 1 + 6 and 3 -+ 2 and does not appear in eqn. (2). By following similar reversible pathways the step corresponding to hydride decomposition is: AG(4 + 5) = id
2
dp + 4’” Vn dp
Peq
(3)
pd
Steps 1 + 2 and 4 + 5 are the only irreversible steps in the cycle so that the free energy dissipated in the cycle is the sum of AG( 1 -+2) and AG(4 --f 5) given by eqn. (4) -J
diG = Jf
J$
dp - Jf VH dp = RT ln[$l”2
pd
- V&f
-pd)
(4)
Pd
The enthalpy and entropy changes can be evaluated in the same manner by following reversible paths. We obtain M(l+
2) =
Peq V 5 s Pf
(1- oH2T) dp - lAI&,I + J'Vn(1 - or,T) dp Peq
(5)
250 Azq4 9 5) = idJ$l Peq
-%$f’)
dp + lMH,,I + /peqVH(l -aHT)
dp
(6)
J'd
where on = VH-1(t3VH/XZ’),. The sum of AH(1 + 2) and AH(4 + 5) is m(l+2)+m(4+5)=--f
~(l-aH,T)dp+yfv~(l--oIHndp pd
(7)
pd
For the entropy changes we obtain p-2
AS(l-+2)=-
~~~,d~-]~pq]-~~oi~~~d~
s
Pf
jd+
Mt4 + 5) = -
(8) peq
CY~,
dp + ]A$,]
- i’”
aHVH dp
(9)
Pd
Peq and ds(1 + 2) + AS(4 + 5) = if 2
CYn,dp - ~~~Hv~ dp
pd
fW
J’d
Generally the entropy change can be written as dS = diS + d,S
(11)
where diS is the entropy produced in the system and d,S is the entropy change in the system as reflected by heat transfer with the surroundings. Forstep1-+2 (12)
AiS(1+ 2) = AS(l + 2) - A,S(l + 2)
where A$(1 -+ 2) = Q(1 + 2)/T = AH( 1 + 2)/T and therefore using eqns. (5) and (8), AiS(l -+ 2) can be written as
A,s(l+2)=j* 52 c&-f*: Peq
2T
(13)
dp
P-J
and similarly AiS(4 + 5) can be obtained from eqns. (6), (9) and (11) Pea
AiS(4 + 5) = $ Pd
Per2
2
dp-$
2
dp
(14)
Pd
and since for the reversible steps 2 + 4 and 5 + 1 AiS is zero, we obtain for the cycle
251
jd,S=[!!$
a-i’:
dp=--_!_J
d,G=iRln
i
-
vH(pFpd)
pd
(15) Although $d$ and $dfG are not zero for the cycle, $ ds, $ dG and $ d.H must, of course, be zero for the complete cycle. This means that _#d,S exactly cancels f diS and the changes AH( 1+ 2) + AH(4 + 5) are exactly canceled by the enthalpy changes which occur during the two reversible steps 2 + 4 and 5 -+ 1. It is of interest to calculate the reversible work done on the system by the surroundings during the complete cycle. $dU’=jf2
(16)
dp-_jfVHdp pd
pd
The work done on the solid cancels in steps 5 + 1 and 2 -+ 4 but not in steps 1 + 2 and 4 * 5. As for the low pressure regime, the net work done on the system is equal to the free energy dissipation in the cycle (eqn. (4)) and the net heat dissipated to the surroundings, i.e. JdQ=-JdW=-J’%
dp+ pd
3. Thermodynamic measurements
JfVndp=JdiG
(17)
Pd
parameters
region derived from p-c-T
for the two-phase
In the low pressure regime thermodynamic parameters can be derived from the temperature dependence of pf and pd but in the high pressure regime these plateau pressures must be converted to fugacities in order to derive thermodynamic parameters. is not known with certainty, it is, none the less, Although P&,) informative to consider the value of the enthalpy change which would be derived from its temperature dependence. Next we will consider the analogous enthalpies derived from pf(ff) and p&d) and their relationships to the enthalpy change derived from p,,(f,,). For hydride formation at peq AG(5 -f 6) = AG,Cp,,)
-
pw
s
= 0 = AH;-
TAS;+
s
vdl-
~HT)
dp
peq
v
T(i-cru,T)dp+TJ
1
oHVHdp 1
peq v -TJ
+Ynldp 1
(18)
253
From eqn. (2), l/2
- V&,
- pes) = -AG(
1 --f 2) = wf
where wp is the plastic, irreversible work done on the solid phase needed to accomodate the volume changes during hydride formation [ 5, 61. The analogous quantity for hydride decomposition is wpd.These irreversible work terms are assumed to be independent of temperature at least over the temperature range where the phase boundary compositions do not change significantly with temperature. From eqn. (25), R 3 In
i
ff”’
aT-’
i
apf
= LWi(app) = -lAWI
+ wg + VHpf + V, T aT-’
This can be expressed in terms of 1kI&(app)l
~Sw)
= -I ~&(wp)l
+4 +
vH
r
using eqn. (22)
ati, -peq) aT-'
+ VH(Pf
Proceeding in a similar way for hydride decomposition, obtain _-R
(26)
-Peq)
(27)
step 4 + 5 we
a In
fd1j2 v, ap, = AKi(wp) = lAIJH”I- V,p, - r aT-’ + wpd i aT-’ I
or expressing this in terms of I AK&(app)l
A-Ki(app)= I~~,(app)I
+ wX +
VH
r
, (eqn. (22)),
abeq-Pd) aT-'
gives
+ VHb-bq
-Pd)
(29)
It should be emphasized that hysteresis affects the derived values of Mi(app) and U”,(app) significantly because at high pressures hysteresis is a large fraction of the AH’ values. It should also be noticed that the determination of M” from AW(app) by eqns. (26) and (29) is not straightforward because the correction terms are comparable with the enthalpy values.
4. High pressure hysteresis for nickel hydride The values of the calorimetric enthalpies determined by Tkacz and Baranowski [2] for nickel hydride have been noted in the introduction. Now the calorimetric enthalpies for hydride formation and decomposition correspond to eqns. (5) and (6) because the heats detected by the calorimeter are equal to the enthalpy changes under constant pressure conditions. Therefore for Ni-H
254
p-l
A&(d)
= AH(1 -+ 2) =
v
J
+ j’V,dp
+(l-on,T)dp-lAZY~ql
JJf
-5.02
=
A&(cal)
= M(4 =
Peq
kJ (mol H)-’
+ 5) = j”? PW
(30)
(1 -cuulT)
dp + ]Lw,,I
+ j”
Vu dp
pd
3.13 kJ (mol H)-’
(31)
where the uncertainty in these quantities is reported to be +O.l kJ (mol H)-r . From these two values we obtain for the sum: AEZ(1 -+2) + M(4 + 5) = -1.89 + 0.2 kJ (mol H)-‘. From eqn. (7) and the known properties of H,(g) [ 3, 41 we obtain
m(l+2)+m(4+5)=-j$l-C$T)dp+@,--p,)Vu Pd
= -1.96
+ 0.48 = -1.48
kJ (mol H)-’
where it has been assumed that VH = 1.7 X lop6m3 (mol H)-’ [8] and independent of pressure. The agreement with experiment is not bad considering that VH may be somewhat smaller at [H]/[Ni] - 1 than 1.7 X 10e6 m3 (mol H)-’ [ 81. Next the free energy dissipation due to hysteresis will be evaluated for a real high pressure hydride - the Ni-H system. From eqn. (4)
-J
2
diG = if
dp - VH(Pf -Pd)
= RT ln(+)“2
- VH(JIf -Pd)
Pd
= 2.71 - 0.48 = 2.23 kJ (mol H)-’ where it has again been assumed that VH = 1.7 X 10e6 m3 (mol H)-’ independent of pressure. The values used for evaluation of $ diG are taken from experimental results [ 23. This hysteresis is greater than twice that observed for Pd-H in the low pressure regime. This value for Ni-H can be compared with a value calculated from eqns. (7) and (lo), i.e. - f diG =-{IvI(l
+ 2) + M(4
+ 5)) - T(AS(1 + 2) + AS(4 + 5))
Pf
=-
s
?(l-
Pd
Pf
%,T)dP+@f-Pd)VH-
V J -+-++P Pd
(32)
255
The term onVn dp has been dropped from aS(l + 2) + AS(4 + 5) because it is small. For the enthalpy terms we take -1.89 + 0.2 kJ (mol H)-’ which is the experimental value from the calorimetric results [2] and the last term can be evaluated from the known properties of H,(g). We obtain -$ diG = +2.62 kJ (mol H)-’ which is somewhat greater than that found from experimental values of pf and pd, but it is probably within experimental error. So far only thermodynamic parameters for the hysteresis cycle have been considered and these are independent of assumptions about the location of equilibrium. Thermodynamic parameters for hydride formation and decomposition will now be evaluated and for this purpose an assumption about the location of equilibrium is needed. We assume that f,,* * fffd which leads to the following for Ni-H (2): pes = 4675 bar (f,, = 8.5 X 104) using the experimental values pi = 6200 bar (ff = 2.43 X 105), pd = 3400 bar (fd = 2.97 X 104). Using these values and eqns. (5) and (6), we obtain Mf(cal)
= -5020
PeqVH
J (mol H)-’ =
s +
from which we obtain IA&,( sition Md(d)
= 3130 J
(mOl
aHIT)
2(l-
Pf (Pf
dp
-
b-&,I
(33)
--Pe.dVH
= 4150 J (mol H)-l and for hydride decompo-
H)-l = j”
‘2
(1 -on,T)
dp + IA&,1
Peq +
be,
(34)
-PdlVH
from which we obtain IMe, = 3970 J (mol H)-‘. The values of 1AI&.,1 should be nearly identical and they are, considering the experimental error of the calorimetric values [2]. If the calorimetric values are as accurate as stated, this good agreement lends support for the assumption that f,,* = fffd. The plastic deformation which occurs for both hydride formation and decomposition is therefore of about equal magnitude so that wp = wpd. The entropy changes M(1 -+ 2) at pf and AS(4 + 5) at pd can be calculated for Ni-H from eqns. (1) and (5) for formation (1 -+ 2) and eqns. (2) and (6) for decomposition (4 + 5). For example, at T = 298 K
As(l+ 2) =
U(l -+2) T
+ $Pf-Peq)
-
AG(lT--, 2) = m$cal)
=-
5020
T
_
+1115= -13.1 T
2
dp
J K-’ (mol H))’
and for aS(4 + 5) = 14.24 J K-’ ( mol H))‘. These magnitudes should differ by
256 pi
s
2
on, dp = 2.5 J K-l (mol H)-’
J’d
and they differ by somewhat less than this but their difference is within experimental error.
5. Correction of thermodynamic parameters to 1 atm Tkacz and Baranowski [ 21 corrected their high pressure enthalpy values to values expected at 1 atm. They employed the equation given in the Introduction to correct formation and decomposition values. It can be seen from eqns. (5) and (6) that effects of hysteresis should disappear in the calorimetric enthalpies, m(l --f 2) and M(4 + 5), as p + 1 atm, if, as assumed, the forward and reverse reactions are identical. We correct the observed values as follows:
= -5020
f 4409 - 1054 = -1665
J (mol H)-l (35)
pd c aH,T)
dp
+ v,Pd
= 3130 - 2460 + 578 = 1248 J (mol El)-’ Thus it can be seen that the magnitudes of the two enthalpies are reasonably close. This supports the present analysis because we have argued that the influence of hysteresis on the enthalpies determined calorimetrically should go to zero as p -+ 1 atm. Tkacz and Baranowski [2] give quite different values for 1 atm conditions than those derived here; they omitted the hydrostatic terms for their correction to 1 atm but there also appears to be an error in the sign of the correction term with respect to the sign of their A& and A& values both of which are reported as positive. It is interesting that the high pressure conditions lead to a more exothermic enthalpy of hydride formation as compared with that at 1 atm; this is one of the factors that allows the hydride phase to form despite the positive mH0, values (see below). It is also of interest to calculate the entropy change expected at 1 atm
= -13.6
- 35.1 = -48.7
J K-’ (mol H)-”
257
This 1 atm calculated value is quite reasonable because, e.g., for Pd-H, AS”,= -46 J K-’ (mol H)-i and it has been noted that there is not much variation amongst aS(; values for various metals. For Ni-H in passing from conditions of 1 atm to peq, AH decreases by 2.6 kJ (mol H)-’ and --T AS decreases by 10.5 kJ (mol H))’ and thus both terms contribute to the formation of the hydride phase at high pressures. As(l+ 2) and A@ 4 + 5) can also be corrected to conditions of 1 atmH,. We have aSi = AS{1 + 2) - If
2
ou, dp = -49.4
J K-’ (mol H)-’
1 &i$
= As(4 -+ 5) -j
?
on, dp = 48.03 J K-i (mol H)-l
pd
These two values differ slightly but their average is close to I AS& I = 48.7 J K-l (mol H))‘. This demonstrates that hysteresis does not affect the standard entropy values much. It should be emphasized, however, that ASi or AS’: (1 atm) should not be calculated from AH( 13 Z)(cal)/!I’ - R In ff”2 or m(4 + 5)(cal)/T + R In fd1’2. These equations give -68.8 J K-’ (mol H))’ and +53.3 J K-’ (mol H)-’ respectively for Ni-H. This is not correct as we have shown before for the low pressure regime 151. Antonov et al. [9] measured pf and pa values for Ni-H over a wide temperature range and these data seem to be compatible with the 298 K data of Tkacz and Baranowski [ 21. In order to be able to employ these data to derive values of ~~(app) from the ~mperature dependence of ff and fd (eqns. (26) and (28)) reliable relationships between f and p must be available for hydrogen above 307 K, which is the limit of the results of Mills et al. [3]. Fukai and Sugimoto [lo] have corrected the results of Mills et al. f33 in order for them to apply at temperatures above 307 K. Using Table 1 in the paper of Fukai and Sugimoto [lo] we have derived from the data of refs. 2 and 9, U”f(app) = -3.1 kJ (mol H)-l and mi(app) = 5.9 kJ (mol H))’ from plots of In ff1’2 and In fd1’2against T-l. From eqns. (26) and (28) and using the values of A@ derived in calculations (35) we obtain (in the order of terms which appear in eqns. (26) and (28)): A.E?i(app) = -116651 oJii(app)
+ 1530 - 1960 + (223012) = -1454
= + I12481 - 1020 + 1884 + (2230/2)
J (mol H)-’
= 3227 J (mol H)-’
The agreement of the calculated values with those determined from the experimental high pressure data is not very good but this may be due largely to the errors in the pressure-fugacity relationships above 307 K. In any case it is seen that in both cases the absolute value of the decomposition enthalpy is greater than the value for hydride formation. In this section we wish to make the point that enthalpies (and entropies) derived from van% Hoff plots for high pressure hydrides do not give the 1 atm values nor the high pressure
258
calorimetric values, and to obtain the 1 atm values from the apparent values requires correction terms which are comparable with the values themselves. 6. Why is hysteresis large for high pressure hydrides? It has been noted by Baranowski and coworkers in their research on high pressure hydrides that hysteresis is invariably large for these systems as compared with low pressure hydrides. No general explanation of this has been given although Baranowski and Bochenska [ll] have discussed the specific case of Ni-H using the model proposed by Scholtus and Hall [12] for hysteresis. It should be pointed out that it has not been possible to predict quantitatively the hysteresis expected for specific low pressure systems so that to expect a quantitative explanation for the high pressure hydrides is unrealistic, however, we will attempt to indicate in a general way why high pressure hydrides have large hysteresis. It should first be pointed out that hysteresis is not given by RT In(ff/ fd)1’2 for high pressure hydrides but the term V&Q -pd) must be subtracted from it (eqn. (4)) to give -$ diG. Often hysteresis in these high pressure hydrides is assumed to be given only by the RT l~~(f~/f~)“~ term but even after V&I, -pd) is subtracted from RT ln(ff/fd)1’2, the values of -9 diG are still quite large for high pressure hydrides and an explanation is provided in the following. Welch and Pick [13] have given a correlation between MU, and @/a where A& is the heat of solution of hydrogen at infinite dilution of hydrogen, p is the shear modulus and Q is the atomic volume of the metal The correlation shows thai MH”, increases with increasing p/a. Amongst these metals the largest variation comes from p rather than C?. It is clear from an examination of the literature that AE?& values for high pressure hydrides are greater than for low pressure hydrides. For example, for Ni-H, A.?&= 27 kJ (mol H)-i [14] and for Pd-H, AZY$ = -10.2 kJ (mol H)-i [15]. According to the regular interstitial solution model [ 161 RT In peq1’2 = AG: = AH& - T AS: + h1/2 where h, is the H-H interaction parameter which is negative and comparable for most systems. This simple model, although not quantitatively satisfactory, suffices to show how the standard free energy for hydride formation depends on A.?& it should be noted that AS; is similar for most systems. Now if LM”, is large, it follows from the correlation between AiYk and or [13] that p is large. Hysteresis in metal hydrides is believed to be due to the irreversible plastic work needed to accomodate the volume changes [6] and the energy needed for this plastic work is expected to be proportional to the energy for dislocation formation which, in turns, is proportional to ,V [ 171. Hence it follows that hysteresis should be large for high pressure hydrides. Let us apply these considerations quantitatively to Ni-H. For Ni-H, -$ diG = 2.23 kJ (mol H)-i [2]. For Pd-H, -9 diG = 0.850 kJ (mol H)-’
259
[6]. The abrupt lattice change is smaller for Pd-H than for Ni-H and therefore hysteresis should be larger for the latter by the factor: (b - U)Ni-n/ (b - c_I)~,_-~= 1.66. The shear modulus of nickel is greater than that for palladium by a factor of 1.47 [18]. These two factors should influence hysteresis multiplicatively and therefore -$ diG should be (1.66) (1.47) (0.850) kJ (mol H)-’ = 2.08 kJ (mol H)-’ in comparison with the experimental value of 2.33 kJ (mol H)-l which is reasonably good agreement. One further example will be considered and this is the PdRh (XRh = 0.80) alloy where the hysteresis is quite large [ 191. For this alloy -$ diG = 4.50 kJ (mol H)-l. Now proceeding in the same way as for Ni-H we have: (1.66) (2.5) (0.85) kJ (mol H)-l = 3.52 kJ (mol H)-’ where the shear modulus has been taken to be equal to the moduli of the pure elements times the mole fraction of each element in the alloy. The agreement is not as good as for the case of Ni-H but the plateau pressure for hydride formation is not as well defined as for the case of Ni-H but a large hysteresis is predicted. We have avoided considering any high pressure hydrides where there is a change of structure on formation of the hydride phase, e.g. Mn-H [20], because hysteresis may be more complex in these cases. For the two hydrides considered above there is merely an abrupt lattice expansion or contraction on hydride formation or decomposition.
7. Conclusions In the present paper we have shown that the entropy production for a high pressure hysteresis cycle is given by eqn. (15). In contrast with conditions of normal pressures, where ideal gas behavior can be assumed and hydrostatic effects on the solid phases are insignificant, the enthalpies for hydride formation and decomposition measured calorimetrically differ significantly at high pressures. The derived equations for high pressure hysteresis are employed for the Ni-H system. An explanation is offered for the fact that hysteresis appears to be large for high pressure hydrides as compared with normal hydrides which form and decompose at pressures below about 100 atm. Acknowledgments TBF and HSC wish to acknowledge National Science Foundation.
the financial
support
of the
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