1.

Rate constant K of a reaction varies with temperature according to the equation log K = constant. \(\frac{E_a}{2.303 R} \frac{1}{T}\)

Answer»

Where Ea is the energy of activation for the reaction. When a graph is plotted for log K versus \(\frac{1}{T}\), a straight line with a slope 6670 K is obtained.

Calculate the energy of activation for this reaction. State units (R = 8.314 JK-1 mol-1)

Slope of the line

\(\frac{-E_a}{2.303 R}\) = -6670 K 

Ea = 2.303 × 8.314 (JK-1 mol-1) × 6670 K 

= 127711.4 J mol-1.



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