Question

Answer the following in brief.

Derive the integrated rate law for the first-order reaction,

\[\ce{A_{(g)} -> B_{(g)} + C_{(g)}}\] in terms of pressure.

Answer 1

i. For the gas phase reaction,

\[\ce{A_{(g)} -> B_{(g)} + C_{(g)}}\]

Let initial pressure of A be P_i that decreases by x within time t.

ii. Pressure of reactant A at time t

P_A = P_i - x  ....(1)

The pressures of the products B and C at time t

P_B = P_C = x

iii. The total pressure at time t is then

P = P_i - x + x + x = P_i + x

Hence, x = P - P_i …(2)

The pressure of A at time t is obtained by substitution of Eq. (1) into Eq. (2).

Thus

P_A = P_i - (P - P_i) = P_i - P + P_i = 2P_i - P

iv. The integrated rate law turns out to be

k = `2.303/"t"  "log"_10 ["A"]_0/["A"]_"t"`

The concentration is now expressed in terms of pressures.

Thus, [A]₀ = P_i and [A]_t = P_A = 2P_i - P

Substitution gives in above

`"k" = 2.303/"t" "log"_10  "P"_"i"/(2"P"_"i" - "P")`  ...(3)

P is the total pressure of the reaction mixture at time t.