JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Chemical Kinetics

For a first order reaction

A (g) \to 2 B (g) + C (g)

at constant volume and 300 K, the total pressure at the beginning

(t = 0)

and at time

t

are

P_{0}

and

P_{t}

, respectively. Initially, only A is present with concentration

[A]_{0}

, and

t_{1/3}

is the time required for the partial pressure of A to reach

1/ 3^{rd}

of its initial value. The correct option(s) is (are)
(Assume that all these gases behave as ideal gases)

Answer: (A), (D)

Solution

For the reaction

A (g) \to 2 B (g) + C (g)

, if pressure

p

of A has reacted, then at time

t

P_{A} = P_{0} - p

P_{B} = 2 p

P_{C} = p

. Total pressure

P_{t} = P_{0} - p + 2 p + p = P_{0} + 2 p

, so

p = (P_{t} - P_{0}) /2

. For first order kinetics:

k t = ln \frac{P _{0}}{P _{0} - p} = ln \frac{2 P _{0}}{3 P _{0} - P _{t}}

. Rearranging:

ln (3 P_{0} - P_{t}) = ln (2 P_{0}) - k t

, which is a linear plot with slope

- k

and intercept

ln (2 P_{0})

. For

t_{1/3}

k t_{1/3} = ln \frac{[ A ] _{0}}{[ A ] _{0} /3} = ln 3

, so

t_{1/3} = \frac{l n 3}{k}

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · One or more correct options