If a reaction follows the Arrhenius equation, the plot $$\ln k$$ vs $$\frac{1}{(RT)}$$ gives straight line with a gradient $$(-y)$$ unit. The energy required to activate the reactant is:

We begin with the Arrhenius equation for the temperature-dependence of the rate constant:

$$k=A\,e^{-E_a/(RT)}$$

Taking natural logarithms on both sides, we use the property $$\ln(e^x)=x$$ to obtain

$$\ln k=\ln A-\dfrac{E_a}{RT}.$$

The question says that a graph of $$\ln k$$ (vertical axis) is drawn against $$\dfrac{1}{RT}$$ (horizontal axis). So, let us write the above equation explicitly in the straight-line form $$y=mx+c$$ with

$$y=\ln k,\qquad x=\dfrac{1}{RT}.$$

Substituting the chosen $$x$$ into the Arrhenius expression, we have

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

Comparing this with $$y=mx+c$$ we identify

$$m=-E_a,$$

because the coefficient of $$x$$ (which is $$1/(RT)$$) is $$-E_a$$, while the intercept $$c$$ is $$\ln A$$.

According to the statement of the problem, the experimentally obtained straight line has a gradient (slope) equal to $$-y$$ unit. Thus,

$$m=-y.$$

But we have already shown that $$m=-E_a$$, so equating the two expressions for the slope gives

$$-E_a=-y\quad\Longrightarrow\quad E_a=y.$$

Therefore, the energy required to activate the reactant, $$E_a,$$ is simply $$y$$ in magnitude and carries the same unit mentioned for the gradient.

Hence, the correct answer is Option B.