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Which of the following correctly represents integrated rate law equation for a first order reaction in a gas phase? - Chemistry

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Question

Which of the following correctly represents integrated rate law equation for a first order reaction in a gas phase?

Options

  • k = `2.303/t xx log_10  P_i/(P_i - P)`

  • k = `2.303/t xx log_10  P_i/(2P_i - P)`

  • k = `2.303/t xx log_10  (2P_i)/(P_i - P)`

  • k = `2.303/t xx log_10  (P_i - P)/(2P_i)`

MCQ
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Solution

`bb(k = 2.303/t xx log_10  P_i/(2P_i - P))`

Explanation:

For the gas phase reaction,

\[\ce{A(g) -> B(g) + C(g)}\]

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

Pressure of reactant A at time t

PA = Pi – x  .....(i)

The pressures of products B and C at time t are

PB = PC = x

The total pressure at time t is then

P = Pi – x + x + x

= Pi + x

Hence, x = P – Pi ........(ii)

Pressure of A, PA at time t is obtained by substitution of equation (ii) into the (i) equation. Thus

PA = Pi – (P – Pi)

= Pi – P + Pi

= 2Pi – P

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 = Pi and [A]t = PA = 2Pi – P

Substitution gives in above

k = `2.303/t log_10  P_i/(2P_i - P)` .....(iii)

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

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