For a reversible reaction R leftharpoons P , at constant...

For the reversible first-order reaction $\text{R} \rightleftharpoons \text{P}$, at equilibrium, the rate of the forward reaction equals the rate of the backward reaction.

$k_f [\text{R}]_{eq} = k_b [\text{P}]_{eq}$

Given $k_b = 4k_f$, we can substitute this into the equilibrium condition:

$k_f [\text{R}]_{eq} = 4k_f [\text{P}]_{eq}$

$[\text{R}]_{eq} = 4[\text{P}]_{eq}$

From the conservation of mass, the total concentration remains constant at any time $t$:

$[\text{R}]_{eq} + [\text{P}]_{eq} = [\text{R}]_0$

Substituting $[\text{R}]_{eq} = 4[\text{P}]_{eq}$ into the mass balance equation gives:

$4[\text{P}]_{eq} + [\text{P}]_{eq} = [\text{R}]_0$

$5[\text{P}]_{eq} = [\text{R}]_0$

$[\text{P}]_{eq} = 0.2[\text{R}]_0$

Using this to find $[\text{R}]_{eq}$:

$[\text{R}]_{eq} = 4 \times 0.2[\text{R}]_0 = 0.8[\text{R}]_0$

Therefore, as $t \to \infty$, the concentration ratio $\dfrac{[\text{P}]}{[\text{R}]_0}$ approaches $0.2$ and $\dfrac{[\text{R}]}{[\text{R}]_0}$ approaches $0.8$. This is correctly represented in the third graph.

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