Question

At 298 K, the rate constant for a reaction is 0.0346 s-1. What is the rate constant at 350K if the Ea = 50.2kJ/mol

Answer 1

Answer:

0.702 /s

Explanation:

Rate constant at $[298 \mathrm{~K}, \mathrm{~K}_{1}=3.46 \times 10^{-2} \mathrm{~s}^{-1}$

Rate constant at $350 \mathrm{~K}, \mathrm{~K}_{2}=?$

$T_{1}=298 \mathrm{~K}$

$T_{2}=350 \mathrm{~K}$

Activation energy, $\mathrm{Ea}=50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}$

Use the following equation to calculate $K_{2} at 350 \mathrm{~K}$

Use the following equation to calculate $K_{2} at 350 \mathrm{~K}$

$\ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]$

Therefore,

 $\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right]$

$\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right]$

$\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=\mathrm{e}^{3.01}$

$\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=20.3$

$K_{2} &=20.3 \times 3.46 \times 10^{-2} \mathrm{~s}^{-1}$

$&=0.702 \mathrm{~s}^{-1}$

hence, the rate constant at $350 \mathrm{~K}$ is 0.702$\mathrm{~s}^{-1}$