Question
What is the activation energy for a reaction whose rate constant doubles when temperature changes from $30^{\circ} C$ to $40^{\circ} C$ ?
✓
Answer
Initial rate constant $= k _1$
and final rate constant $= k _2 ; \frac{k_2}{k_1}=2$
Initial temperature $=T_1=273+30=303 K$
Final temperature $=T_2=273+40=313 K$
Energy of activation $= E _{ a }=?$
$ \log _{10} \frac{k_2}{k_1}=\frac{E_{ a }\left(T_2-T_1\right)}{2.303 R \times T_1 \times T_2}$
$ \therefore E_{ a } =\frac{2.303 R \times T_1 \times T_2}{\left(T_2-T_1\right)} \log _{10} \frac{k_2}{k_1}$
$ =\frac{2.303 \times 8.314 \times 303 \times 313}{(313-303)} \log _{10} 2$
$ =\frac{2.303 \times 8.314 \times 303 \times 313 \times 0.3010}{10}$
$ =54660\ J\ mol ^{-1}$
$ =54.66\ kJ\ mol ^{-1} $
Activation energy $= E _{ a }=54.66\ kj\ mol ^{-1}$
and final rate constant $= k _2 ; \frac{k_2}{k_1}=2$
Initial temperature $=T_1=273+30=303 K$
Final temperature $=T_2=273+40=313 K$
Energy of activation $= E _{ a }=?$
$ \log _{10} \frac{k_2}{k_1}=\frac{E_{ a }\left(T_2-T_1\right)}{2.303 R \times T_1 \times T_2}$
$ \therefore E_{ a } =\frac{2.303 R \times T_1 \times T_2}{\left(T_2-T_1\right)} \log _{10} \frac{k_2}{k_1}$
$ =\frac{2.303 \times 8.314 \times 303 \times 313}{(313-303)} \log _{10} 2$
$ =\frac{2.303 \times 8.314 \times 303 \times 313 \times 0.3010}{10}$
$ =54660\ J\ mol ^{-1}$
$ =54.66\ kJ\ mol ^{-1} $
Activation energy $= E _{ a }=54.66\ kj\ mol ^{-1}$
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