A reactant (A) forms two products:A⟶k1B;     Activation energy EalA→k2 C Activation energy Ea2If Ea/2=2Ea1, then k1 and k2 are related as

# A reactant (A) forms two products:$\mathrm{A}\stackrel{{k}_{1}}{⟶}\mathrm{B};$     Activation energy ${E}_{\mathrm{al}}$ Activation energy ${E}_{\mathrm{a}2}$If ${E}_{\mathrm{a}/2}=2{E}_{\mathrm{a}1}$, then ${k}_{1}$ and ${k}_{2}$ are related as

1. A

${k}_{1}=2{k}_{2}{\mathrm{e}}^{{E}_{\mathrm{a}2}/RT}$

2. B

${k}_{1}={k}_{2}{\mathrm{e}}^{{E}_{a1}/RT}$

3. C

${k}_{2}={k}_{1}{\mathrm{e}}^{{E}_{\mathrm{a}2}/RT}$

4. D

${k}_{1}=A{k}_{2}{\mathrm{e}}^{{E}_{\mathrm{a}1}/RT}$

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### Solution:

We have ${k}_{1}={A}_{1}{\mathrm{e}}^{-{E}_{\mathrm{al}}/RT}$ and ${k}_{2}={A}_{2}{\mathrm{e}}^{-{E}_{\mathrm{a}2}/RT}={A}_{2}{\mathrm{e}}^{-2{E}_{\mathrm{a}1}/RT}$

Hence  $\frac{{k}_{1}}{{k}_{2}}=\frac{{A}_{1}}{{A}_{2}}\frac{{\mathrm{e}}^{-{E}_{\mathrm{al}}/RT}}{{\mathrm{e}}^{-2{E}_{\mathrm{a}1}/RT}}=\frac{{A}_{1}}{{A}_{2}}{\mathrm{e}}^{{E}_{\mathrm{a}1}/RT}$or ${k}_{1}=\left(\frac{{A}_{1}}{{A}_{2}}\right){k}_{2}{\mathrm{e}}^{{E}_{\mathrm{a}\mid }/RT}=A{k}_{2}{\mathrm{e}}^{{E}_{\mathrm{al}}/RT}$

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