The rate constant for the first order decomposition of {\N}_2 {\O}_5 is given by the following equation : k= (2.5 \times 10^{14} { s}^{-1}) e^{(-25000 {\K}) \/ {\T}} Calculate {\E}_a for this reaction and rate constant if its half-life period be 300 minutes.

CBSE Class 12 Chemistry 2020 Outside Delhi Set 1 Solved Paper

Question : 29 of 37

Marks: +1, -0

The rate constant for the first order decomposition of

{\N}_2 {\O}_5

is given by the following equation :

k= (2.5 × 10^{14} {\ s}^{-1}) e^{(-25000 {\K}) \/ {\T}}

Calculate

{\E}_a

for this reaction and rate constant if its half-life period be 300 minutes.

Solution:

\;k= (2.5 × 10^{14} l^{-1}) {\e}^{(-25000 K) \/ {\T}}

\;t_{1 \/ 2}=300 \;\text " minutes "\;

\;\;{E_a}/{R}=25000 {\K}

\;E_a=25000 × {\R} × {\K}

\;=25000 × 8.314 {\J} {\K}^{-1} {\mol}^{-1} × {\K}

\;=207850 {\J} {\mol}^{-1}=207.850 {\KJ} {\mol}^{-1}

\;t_{1 \/ 2}=\;{0.693}/{K} ⇒ {\K}=\;{0.693}/{300} {\min}^{-1}

\;=0.231 × 10^{-2}

\;=2.31 × 10^{-3} {\min}^{-1}

\;

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