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JEE Advanced levelSingle correct optionDifficulty 3 of 3ck025

A gas-phase reaction obeys $- \frac{d c _{A}}{d t} = k_{c} c_{A}^{2}$ with $k_{c} = 0.5 L mo l^{- 1} s^{- 1}$ at $300 K$ . Treat the gas as ideal ( $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ ). If the same rate is written as $- d p_{A} / d t = k_{p} p_{A}^{2}$ , the value of $k_{p}$ is

A $12.3 at m^{- 1} s^{- 1}$
B $0.50 at m^{- 1} s^{- 1}$
C $0.0203 at m^{- 1} s^{- 1}$
D $0.00082 at m^{- 1} s^{- 1}$

Answer: C

With

p_{A} = c_{A} RT

,

- d p_{A} / d t = RT (- d c_{A} / d t) = RT k_{c} (p_{A} / RT)^{2} = (k_{c} / RT) p_{A}^{2}

. So

k_{p} = k_{c} / RT = 0.5/ (0.0821 \times 300) = 0.0203 at m^{- 1} s^{- 1}

.

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