How much faster would a reaction proceed at 2 | vedclass

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(C) Using the Arrhenius equation: $\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} \left[ \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right]$Given: $E_{a} = 6

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) Using the Arrhenius equation: $\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} \left[ \frac{T_{2} - T_{1}}{T_{1} T_{2}} ight]$Given: $E_{a} = 65,000 \, J/mol$,$R = 8.314 \, J/K \cdot mol$,$T_{1} = 273 \, K$,$T_{2} = 298 \, K$.Substituting the values:$\log \frac{K_{2}}{K_{1}} = \frac{65000}{2.303 	imes 8.314} \left[ \frac{298 - 273}{273 	imes 298} ight]$$\log \frac{K_{2}}{K_{1}} = \frac{65000}{19.147} 	imes \frac{25}{81354}$$\log \frac{K_{2}}{K_{1}} = 3394.78 	imes 0.000307 \approx 1.043$$\frac{K_{2}}{K_{1}} = 10^{1.043} \approx 11$Thus,the reaction proceeds approximately $11$ times faster.

How much faster would a reaction proceed at $25\,^{\circ}C$ than at $0\,^{\circ}C$ if the activation energy is $65\,kJ/mol$?

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