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EEJ MAIN Chemistry QUESTION #5535
Question 1
For a reaction $aA+bB\to cC+dD$, the plot of $\log k$ vs $\frac{1}{T}$ is a straight line with slope $= -10000\ K$. Given that the rate constant is $10^{-5}\ s^{-1}$ at 500 K, find the temperature (in K) at which the rate constant equals $10^{-4}\ s^{-1}$. (Round to nearest integer)
  • 526 K
  • 556 K✔️
  • 600 K
  • 1000 K
Correct Answer Explanation
Using the Arrhenius equation in log form: $\log\frac{k_2}{k_1} = \frac{-\text{slope}}{2.303}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\cdot2.303$. Since slope of $\log k$ vs $\frac{1}{T}$ is $-\frac{E_a}{2.303R} = -10000$, then $E_a = 10000\times2.303\times8.314$. Using $\log\frac{10^{-4}}{10^{-5}} = -10000\left(\frac{1}{T_2}-\frac{1}{500}\right)$: $1 = -10000\left(\frac{1}{T_2}-0.002\right) \Rightarrow \frac{1}{T_2} = 0.002-0.0001 = 0.0018 \Rightarrow T_2 \approx 556\ K$.