A reactant (A) forms two products: \(\mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B}\); Activation…
A reactant (A) forms two products:
\(\mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B}\); Activation energy \(E_{\mathrm{a} 1}\)
\(\mathrm{A} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}\); 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
\(k_{1}=2 k_{2} \mathrm{e}^{E_{\mathrm{a} 2} / R T}\)
\(k_{1}=k_{2} \mathrm{e}^{E_{\mathrm{a} 1} / R T}\)
\(k_{2}=k_{1} \mathrm{e}^{E_{\mathrm{a} 2} / R T}\)
\(k_{1}=A k_{2} \mathrm{e}^{E_{\mathrm{a} 1} / R T}\)
Solution
We have \(k_{1}=A_{1} \mathrm{e}^{-E_{\mathrm{a} 1} / R T}\) and \(k_{2}=A_{2} \mathrm{e}^{-E_{\mathrm{a} 2} / R T}=A_{2} \mathrm{e}^{-2 E_{\mathrm{a} 1} / R T}\)
Hence \(\frac{k_{1}}{k_{2}}=\frac{A_{1}}{A_{2}} \frac{\mathrm{e}^{-E_{\mathrm{al}} / R T}}{\mathrm{e}^{-2 E_{\mathrm{a} 1} / R T}}=\frac{A_{1}}{A_{2}} \mathrm{e}^{E_{\mathrm{al}} / R T}\) or \(k_{1}=\left(\frac{A_{1}}{A_{2}}ight) k_{2} \mathrm{e}^{E_{\mathrm{a} 1} / R T}=A k_{2} \mathrm{e}^{E_{\mathrm{al}} / R T}\)