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Chemistry Question on Chemical Kinetics

The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction (R=8.314JK−1 mol−1) ( R = 8.314 JK^{ - 1} \, mol^{ - 1} )

A

41.7 KJ mol−141.7 \, KJ \, mol^{ - 1}

B

4.17 kJ mol−14.17 \, kJ \, mol^{ - 1}

C

417 kJ mol−1417 \, kJ \, mol^{ - 1}

D

0.417 kJ mol−10.417 \, kJ \, mol^{ - 1}

Answer

41.7 KJ mol−141.7 \, KJ \, mol^{ - 1}

Explanation

Solution

log k2k1=E−a2.303 R[T2−T1T1−T2]\frac{ k_2 }{ k_1 } = \frac{ E-a }{ 2.303 \, R } \bigg [ \frac{ T_2 - T_1 }{ T_1 - T_2 } \bigg ]
k2k1=6, T1=350 K, T2=400K\frac{ k_2 }{ k_1} = 6, \, T_1 = 350 \, K, \, T_2 = 400 K
log 6 =Ea2.303×8.314[400−350400×350]= \frac{ E_a }{ 2.303 \times 8.314 } \bigg [ \frac{ 400 - 350 }{ 400 \times 350 } \bigg ]
0.7782 = Ea2.303×8.314×50400×350\frac{ E_a }{ 2.303 \times 8.314 } \times \frac{ 50 }{ 400 \times 350 }
Ea=41.721 J mol−1E_a = 41.721 \, J \, mol^{ - 1}
= 41.721 kJ mol−1mol^{ - 1}