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Question

A first-order reaction is $$50\%$$ completed in $$40$$ minutes at $$300\ K$$ and in $$20$$ minutes at $$320\ K$$ . Calculate the activation energy of the reaction. 

[Given: log $$2=0.3010$$, log $$4=0.6021$$, R=$$8.314$$ $$JK^{-1}$$ $$mol^{-1}$$]


Solution

When a first order reaction is $$50\%$$ completed, the time is equal to half life period.

A first order reaction is $$50\%$$ completed in $$40$$ minutes at $$300$$K and in $$20$$ minutes at $$320$$ K.
$$t_{1/2}=40$$ minutes
$$t'_{1/2}=20$$ minutes

The Arrhenius equation and temperature variation is given by the expression.
$$log\left( \dfrac {k'}{k} \right)= \dfrac {E_a}{2.303R}[\dfrac {T'-T}{TT'}]$$

Also, $$t_{1/2}=\dfrac {0.693}{k}$$ or  $$t_{1/2} \propto \dfrac {1}{k}$$ 

Hence, $$log \left(\dfrac {t_{1/2}}{t_{1/2}'}\right) = \dfrac {E_a}{2.303R}[\dfrac {T'-T}{TT'}]$$

$$log\left( \dfrac {40}{20} \right)= \dfrac {E_a}{2.303 \times 8.314 J/mol/K}[\dfrac {320K-300K}{300K \times 320K}]$$

$$ 0.3010 = \dfrac {E_a}{19.147 J/mol/K}[ 0.0002083 /K]$$

$$E_a=27664 J/mol$$

Chemistry

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