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Pressure of First Order Reaction

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The decomposition of phosphine is irreversible and first order at 650^{o}C.

4PH_{3}(g) -> P_{4}(g) + 6H_{2}(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 ^{o}K.

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 650^{o}C.

Calculations:

Total pressure of the system (π) and partial pressure of A (p_{A}) are related as:

p_{A} = p_{Ao} - (a/Δn)(π - π_{o})

If phosphine is taken as A, then

p_{Ao} = π_{o} = 1 atm

a = stoichiometric coefficient of A = 4

Δn = stoichiometric coefficient of products - stoichiometric coefficient of reactants
= 1 + 6 - 4 = 3

Therefore,

p_{A} = 1 - 4/3 (π - 1),

4/3 (π - 1) = 1 - p_{A}

i.e.,

π = 0.75 (1 - p_{A}) + 1

Rate constant k, at 650^{o}C (= 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 (C_{A}/C_{Ao}) = k t

or

-ln (p_{A}/p_{Ao}) = kt

p_{A}/p_{Ao} = e^{-kt}

p_{A} = p_{Ao} e^{-kt} = 1 x exp(-3.04 x 10^{-3} t) = exp(-3.04 x 10^{-3} t)

for various time values, p_{A} is calculated and tabulated as follows:

t, sec 50 100 500
p_{A}, atm 0.860 0.738 0.219

From the relationship of total pressure(π) and partial pressure of A (p_{A}),

π = 0.75 (1 - p_{A}) + 1

total pressure for various time values are calculated as:

t, sec 50 100 500
p_{A}, atm 0.860 0.738 0.219
**π, atm 1.105 1.197 1.586**

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Last Modified on: 04-Feb-2022

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