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This can be seen in the Arrhenius equation: ln(k)=-Ea/RT+ln(A) where Ea is the activation energy. The slope of the line plotted on a ln(k) vs 1/T graph is -Ea/R. This means that for higher activation energies, the slope is more steeply negative so the ln(k) value, and also the k value, changes more rapidly with time than for lower activation energies.
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