The given figure displays the Arrhenius plots ( k vs (1)/(T) ) for...

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 the y-axis is $\ln k$ and the x-axis is $\dfrac{1}{T}$.

The slope of the line is given by $m = -\dfrac{E_a}{R}$.

From the given graph, the line for Reaction X is steeper (more negative) than the line for Reaction Y. This means the magnitude of the slope for Reaction X is greater than that for Reaction Y.

$\left| m_X \right| > \left| m_Y \right|$

$\dfrac{E_{a,X}}{R} > \dfrac{E_{a,Y}}{R}$

$E_{a,X} > E_{a,Y}$

Therefore, the activation energy of Reaction X is greater than the activation energy of Reaction Y.

Answer: $E_a$ of Reaction X > $E_a$ of Reaction Y