For a reversible reaction $$\mathbf{R}\rightleftharpoons\mathbf{P}$$, at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constants $$k_f$$ and $$k_b$$, respectively. At time zero, the concentration of $$\mathbf{R}$$ is $$[\mathbf{R}]_0$$ and the concentration of $$\mathbf{P}$$ is zero. At any given time, $$[\mathbf{R}]$$ and $$[\mathbf{P}]$$ are the concentrations of $$\mathbf{R}$$ and $$\mathbf{P}$$, respectively. If $$k_b=4k_f$$, the correct graphical representation of the reaction is

Let the initial concentration of the reactant be $$[\mathbf R]_0$$. At any time $$t$$, let $$x$$ be the amount of $$\mathbf R$$ that has been converted to $$\mathbf P$$.

Then $$[\mathbf R]=[\mathbf R]_0-x,$$ $$[\mathbf P]=x.$$

Because both the forward and backward steps are elementary first-order, the net rate of formation of $$\mathbf P$$ is

$$\frac{dx}{dt}=k_f([\mathbf R]_0-x)-k_bx$$ $$\Longrightarrow\frac{dx}{dt}+(k_f+k_b)x=k_f[\mathbf R]_0\; -(1)$$

This is a linear first-order differential equation whose solution is obtained using the integrating-factor method:

$$x=\frac{k_f[\mathbf R]_0}{k_f+k_b}\Bigl(1-e^{-(k_f+k_b)t}\Bigr)\; -(2)$$

Hence the time-dependent concentrations are

$$[\mathbf R]=[\mathbf R]_0-x =[\mathbf R]_0\left[\frac{k_b}{k_f+k_b}+\frac{k_f}{k_f+k_b}\,e^{-(k_f+k_b)t}\right]\; -(3)$$

$$[\mathbf P]=x =[\mathbf R]_0\frac{k_f}{k_f+k_b}\Bigl(1-e^{-(k_f+k_b)t}\Bigr)\; -(4)$$

At equilibrium $$t\to\infty$$, the exponential terms vanish, giving

$$[\mathbf R]_{eq}= [\mathbf R]_0\frac{k_b}{k_f+k_b},\qquad [\mathbf P]_{eq}= [\mathbf R]_0\frac{k_f}{k_f+k_b}.$$

For the data in the problem, $$k_b=4k_f$$, so

$$[\mathbf R]_{eq}= [\mathbf R]_0\frac{4k_f}{5k_f}=0.8[\mathbf R]_0,$$ $$[\mathbf P]_{eq}= [\mathbf R]_0\frac{k_f}{5k_f}=0.2[\mathbf R]_0.$$

Therefore the concentration-time curves must exhibit the following features:

• $$[\mathbf R]$$ falls exponentially from $$[\mathbf R]_0$$ to $$0.8[\mathbf R]_0$$; the drop is only 20 % of the initial value.
• $$[\mathbf P]$$ rises from 0 to $$0.2[\mathbf R]_0$$ with the same time constant.
• The two curves never cross, because $$[\mathbf R]$$ always exceeds $$[\mathbf P]$$ (ratio 4 : 1 at equilibrium).
• Both curves approach horizontal asymptotes as $$t\to\infty$$ (equilibrium).

Among the given diagrams, the only one that shows • $$[\mathbf R]$$ starting at $$[\mathbf R]_0$$ and leveling off near 0.8$$[\mathbf R]_0$$, • $$[\mathbf P]$$ starting at 0 and leveling off near 0.2$$[\mathbf R]_0$$, • no intersection of the two curves, is Option C.

Option C which is: the correct graph.