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Question: The activation energies of two reactions are \(E_{a_{1}}\) and \(E_{a_{2}}\)with \(E_{a_{1}} > E_{a_...

The activation energies of two reactions are Ea1E_{a_{1}} and Ea2E_{a_{2}}with Ea1>Ea2E_{a_{1}} > E_{a_{2}} If the temperature of the reacting systems is increased from T to T´, which of the following is correct?

A

K1′K1\frac{K_{1}^{'}}{K_{1}} = K2′K2\frac{K_{2}^{'}}{K_{2}}

B

K1′K1\frac{K_{1}^{'}}{K_{1}}< 2 K2′K2\frac{K_{2}^{'}}{K_{2}}

C

K1′K1\frac{K_{1}^{'}}{K_{1}}>K2′K2\frac{K_{2}^{'}}{K_{2}}

D

K1′K1\frac{K_{1}^{'}}{K_{1}}<K2′K2\frac{K_{2}^{'}}{K_{2}}

Answer

K1′K1\frac{K_{1}^{'}}{K_{1}}>K2′K2\frac{K_{2}^{'}}{K_{2}}

Explanation

Solution

ln K´K\frac{K´}{K}=(1T−1T´)\left( \frac{1}{T} - \frac{1}{T´} \right)= EaR\frac{Ea}{R}×ΔTTT′\frac{\Delta T}{TT'}

DT is positive, therefore greater the Ea value, greater will be value of K´K\frac{K´}{K}

\ K1´K1\frac{K_{1}^{´}}{K_{1}} >  K2´K2\frac{K_{2}^{´}}{K_{2}}