The decomposition of phosphine is irreversible and first order at 650oC.
4PH3(g) -> P4(g) + 6H2(g)
The rate constant (sec-1) is reported as
log k = (-18993/T ) + 2 log T + 12.13 (log to the base 10)
where T is in oK.
In a closed vessel (constant volume) initially containing phosphine at 1 atm pressure, what will be the pressure after 50, 100 and 500 sec. The temperature is maintained at 650oC.
Calculations:
Total pressure of the system (π) and partial pressure of A (pA) are related as:
pA = pAo - (a/Δn)(π - πo)
If phosphine is taken as A, then
pAo = πo = 1 atm
a = stoichiometric coefficient of A = 4
Δn = stoichiometric coefficient of products - stoichiometric coefficient of reactants= 1 + 6 - 4 = 3
Therefore,
pA = 1 - 4/3 (
π - 1),4/3 (
π - 1) = 1 - pAi.e.,
π = 0.75 (1 - pA) + 1
Rate constant k, at 650oC (= 650 + 273 = 923 K) is found from the relationship for k and T
log k = (-18993/T ) + 2 log T + 12.13
log k = -18993/923 + 2 log 923 + 12.13
log k = -2.517
k = 10-2.517
Therefore,
k = 3.04 x 10-3 sec-1.For the first order equation,
-ln (CA/CAo) = k t
or
-ln (pA/pAo) = kt
pA/pAo = e-kt
pA = pAo e-kt = 1 x exp(-3.04 x 10-3 t) = exp(-3.04 x 10-3 t)
for various time values, pA is calculated and tabulated as follows:
t, sec 50 100 500 pA, atm 0.860 0.738 0.219
From the relationship of total pressure(
π) and partial pressure of A (pA),π = 0.75 (1 - pA) + 1
total pressure for various time values are calculated as:
t, sec 50 100 500 pA, atm 0.860 0.738 0.219 π, atm 1.105 1.197 1.586
Last Modified on: 04-Feb-2022
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