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

0.702 /s

Step-by-step explanation:

Rate constant at
`[298 \mathrm{~K}, \mathrm{~K}_(1)=3.46 * 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 * 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 * 10^(-2) \mathrm{~s}^(-1)}\right) &=\frac{50.2 * 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 * 10^(-2) \mathrm{~s}^(-1)}\right) &=\frac{50.2 * 10^(3) \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^(-1) \mathrm{~mole}^(-1)}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} * 350 \mathrm{~K}}\right]`

`\frac{K_(2)}{3.46 * 10^(-2) \mathrm{~s}^(-1)} &=\mathrm{e}^(3.01)`

`\frac{K_(2)}{3.46 * 10^(-2) \mathrm{~s}^(-1)} &=20.3`

`K_(2) &=20.3 * 3.46 * 10^(-2) \mathrm{~s}^(-1)`

`&=0.702 \mathrm{~s}^(-1)`

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