1.

Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation  log k = log A   \(\frac{E_a}{2.303R}\) \(\Big(\frac{1}{T}\Big)\)Where Ea is the activation energy. When a graph is plotted for logk Vs \(\frac{1}{T}\) a straight line with a slope of- 4000K Is obtained. Calculate the activation energy.

Answer»

log k = log A - \(\frac{E_a}{2.303R}\) \(\Big(\frac{1}{T}\Big)\)

y = c + mx

m =  -\(\frac{E_a}{2.303R}\)

Ea = -2.303Rm 

Ea = – 2.303 x 8.314 J K-1 mol-1 x (-4000 K)

Ea = 76,589 J mol-1

Ea =76589kJmol-1



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