# Total Pressure

The decomposition of phosphine is irreversible and first order at 650oC.

4PH3(g) à 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 (p) and partial pressure of A (pA)are related as:

pA = pAo - (a/Dn)(p - po)

If phosphine is taken as A, then

pAo = po = 1 atm

a = stoichiometric coefficient of A = 4

Dn = stoichiometric coefficient of products - stoichiometric coefficient of reactants

= 1 + 6 - 4 = 3

Therefore,

pA = 1 - 4/3 (p - 1),

4/3 (p - 1) = 1 - pA

i.e.,

p = 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(p) and partial pressure of A (pA),

p = 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
p, atm	1.105	1.197	1.586```