For a reversible reaction R ⇌ 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 R is [R]₀ and the concentration of P is zero. At any given time, [R] and [P] are the concentrations of R and P, respectively. If k_b = 4k_f, the correct graphical representation of the reaction is:

The correct answer is Option C, showing [R]/[R]₀ leveling off at approximately 0.8 and [P]/[R]₀ leveling off at approximately 0.2 at equilibrium.

Step 1: Set Up the Rate Equation

For the reversible elementary reaction R ⇌ P, both forward and backward steps are first order. The net rate of change of [R] is:

$\frac{d\left[R\right]}{dt}=-k_f\left[R\right]+k_b\left[P\right]$

Since mass is conserved, at any time: [P] = [R]₀ - [R]. Substituting:

$\frac{d\left[R\right]}{dt}=-k_f\left[R\right]+k_b({\left[}_R{}_0-[R\left]\right)$

$\frac{d\left[R\right]}{dt}=-(k_f+k_b)\left[R\right]+k_b{\left[}_R$

Step 2: Find the Equilibrium Concentrations

At equilibrium, d[R]/dt = 0:

$0=-(k_f+k_b){\left[}_R+k_b{\left[}_R$

${\left[}_R=\frac{k_b}{k_f+k_b}{\left[}_R$

Now substitute the given condition k_b = 4k_f:

${\left[}_R=\frac{4k_f}{k_f+4k_f}{\left[}_R=\frac{4}{5}{\left[}_R=0.8{\left[}_R$

And therefore:

${\left[}_P={\left[}_R-{\left[}_R=0.2{\left[}_R$

Step 3: Verify the Direction of Equilibrium

The equilibrium constant K_eq for this reaction is:

$K_{eq}=\frac{k_f}{k_b}=\frac{k_f}{4k_f}=\frac{1}{4}=0.25$

Since K_eq < 1, equilibrium strongly favors the reactant R. This means R is not fully consumed — most of it remains at equilibrium, while very little P is formed. This is the key physical insight.

Step 4: Analyze the Shape of the Curves

The general solution to the differential equation gives exponentially approaching curves:

$\frac{\left[R\right]}{{\left[}_R}=0.8+0.2e^{-(k_f+k_b)t}$

$\frac{\left[P\right]}{{\left[}_R}=0.2-0.2e^{-(k_f+k_b)t}$

• At t = 0: [R]/[R]₀ = 0.8 + 0.2 = 1 ✓, and [P]/[R]₀ = 0.2 - 0.2 = 0 ✓
• As t → ∞: [R]/[R]₀ → 0.8 and [P]/[R]₀ → 0.2
• Both curves approach their equilibrium values exponentially (not linearly)

Step 5: Match to the Correct Graph

Looking at the four option images:

• Option A shows both [R] and [P] converging near 0.5 — this would be the case only if k_f = k_b. Incorrect.
• Option B shows [P] reaching ~0.8 and [R] dropping to ~0.2 — this would require k_f = 4k_b, the opposite condition. Incorrect.
• Option C shows [R]/[R]₀ (dashed) leveling off at ~0.8 and [P]/[R]₀ (solid) leveling off at ~0.2, with smooth exponential curvature. This perfectly matches our calculated equilibrium values. Correct!
• Option D shows linear curves — impossible for first-order kinetics with exponential approach to equilibrium. Incorrect.

Conclusion: Since k_b = 4k_f, the backward reaction is 4 times faster than the forward reaction, so the equilibrium strongly favors the reactant R. At equilibrium, 80% of the original reactant remains as R and only 20% is converted to P. The graph in Option C, showing [R]/[R]₀ → 0.8 and [P]/[R]₀ → 0.2 with smooth exponential curves, is the only correct representation.