The given plot shows the variation of k versus 1 T for two different chemical reactions, Reaction 1 and Reaction 2. Based on the graph, which of the following relations is correct regarding their activation energies ( E_a ) and pre-exponential factors ( A )?

Step-by-step solution

According to the Arrhenius equation: k = A e^ -E_a / RT Taking the natural logarithm on both sides, we get: k = A - E_a R ( 1 T ) This represents a straight line equation y = mx + c , where y = k and x = 1 T . The slope of the line is m = - E_a R and the y-intercept is c = A . From the given graph, both Reaction 1 and Reaction 2 have the same y-intercept. Therefore: A₁ = A₂ A₁ = A₂ The graph also shows that the line for Reaction 1 is steeper (has a more negative slope) than the line for Reaction 2. Let m₁ and m₂ be