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  • The rate constant <img alt="\mathrm{\mathrm{K}_1 \text{ and } \mathrm{K}_2}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cmathrm%7BK%7D_1%20%5Ctext%7B%20and%20%7D%20%5Cmathrm

The rate constant \mathrm{\mathrm{K}_1 \text{ and } \mathrm{K}_2} for two different reactions are 10^{11} \cdot \mathrm{e}^{-3000} / \mathrm{T} \text{ and } 10^{10} \cdot \mathrm{e}^{-1500} / \mathrm{T}, respectively. The temperature at which \mathrm{k_{1}} =\mathrm{k_{2}} is:

Option: 1

1000 K


Option: 2

1500/2.303 K


Option: 3

2000 K


Option: 4

1000/2.303 K


Answers (1)

best_answer

Given rate constant 

\begin{aligned} & \mathrm{K_1}=10^{11} \mathrm{e}^{-3000} / \mathrm{T} \\ \\& \mathrm{K_2}=10^{10} \mathrm{e}^{-1500} / \mathrm{T} \end{aligned}

\begin{aligned} \because \quad \mathrm{k_1=k_2} \Rightarrow & \mathrm{10^{11} e^{-3000 / T}=10^{10} e^{-\frac{1500}{T}}} \\ \\& \mathrm{10 e^{-3000 / T}=e^{-\frac{1500}{T}}} \end{aligned}

By Applying in both side 

\begin{aligned} & \ln \left(10 e^{-3000 / T}\right)=\ln \left(e^{-\frac{1500}{T}}\right) \\ \\& \mathrm{ \ln (10)-\frac{3000}{T}=\frac{-1500}{T}} \\ \\& \mathrm{2.303 \log 10=\frac{1500}{T}} \end{aligned}

\mathrm{T=\frac{1000}{2.303} \mathrm{~K}}

Posted by

Devendra Khairwa

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