The slope of the line for the graph of log k versus 1T for the reaction, N2O5→2NO2+12O2 is -5000. Calculate the energy of activation of the reaction (kJ K−1 mol−1)

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95.7

9.57

957

0.957

answer is A.

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Detailed Solution

K=Ae−Ea/RTln⁡ K=ln⁡A−EaRT2.303log⁡K=−EaRT+2.303log⁡Alog⁡K=−Ea2.303RT+log⁡ASlope =-Ea2.303R=-5000Ea=5000×8.31×2.303×10-3=95.7 kJ k−1 mol−1

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