The given plot shows the variation of k versus (1)/(T) for two...

According to the Arrhenius equation:

$k = A e^{-E_a / RT}$

Taking the natural logarithm on both sides, we get:

$\ln k = \ln A - \dfrac{E_a}{R} \left(\dfrac{1}{T}\right)$

This represents a straight line equation $y = mx + c$, where $y = \ln k$ and $x = \dfrac{1}{T}$.
The slope of the line is $m = -\dfrac{E_a}{R}$ and the y-intercept is $c = \ln A$.

From the given graph, both Reaction 1 and Reaction 2 have the same y-intercept. Therefore:

$\ln A_1 = \ln A_2 \Rightarrow A_1 = A_2$

The graph also shows that the line for Reaction 1 is steeper (has a more negative slope) than the line for Reaction 2. Let $m_1$ and $m_2$ be their respective slopes.

$m_1 < m_2$

$-\dfrac{E_{a1}}{R} < -\dfrac{E_{a2}}{R}$

Multiplying by $-R$ (which reverses the inequality):

$E_{a1} > E_{a2}$

Thus, $E_{a1} > E_{a2}$ and $A_1 = A_2$.

Answer: $E_{a1} > E_{a2}$ and $A_1 = A_2$