For a reversible reaction R ⇌ P, at constant temperature, both the forward and the backward reaction

Q: Why do we use $[R]_0 = [R]_{eq} + [P]_{eq}$?

Since the reaction starts with only R and no P, every molecule of P formed must come from a molecule of R. Therefore, the total number of moles (and thus concentration, since volume is constant) of R and P at any time must equal the initial concentration of R.

Q: How do we know the curves are exponential?

For first-order reversible reactions, the integration of the rate law yields an exponential approach to equilibrium. The concentration changes as $e^{-(k_f + k_b)t}$.

Q: What if $k_f$ was equal to $k_b$?

If $k_f = k_b$, then $K_c = 1$, meaning $[P]_{eq} = [R]_{eq} = 0.5[R]_0$. Both curves would meet and settle at the 0.5 mark, which corresponds to option (A).

Q: Why doesn't the concentration of R drop to zero?

Because it's a reversible reaction! As soon as some P is formed, the backward reaction starts converting P back into R. Equilibrium is reached when these two opposing rates balance each other out, leaving a non-zero amount of R.

Question

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

Sigma Percentile · Accepted Answer

### The Beauty of Reversible Reactions

Imagine a busy two-way street. Cars are moving forward, and cars are moving backward. In the world of chemical kinetics, this is exactly what a reversible reaction looks like. We have a reactant $	ext{R}$ converting into a product $	ext{P}$ at a forward rate constant $k_f$, while simultaneously, $	ext{P}$ is converting back into $	ext{R}$ at a backward rate constant $k_b$.

The question asks us to visualize this dynamic dance over time. We start with a full tank of $	ext{R}$ (concentration $[	ext{R}]_0$) and absolutely zero $	ext{P}$. As time ticks forward, $	ext{R}$ gets consumed and $	ext{P}$ is formed. But because it's a reversible reaction, $	ext{R}$ will never completely disappear. Eventually, the system reaches a state of **dynamic equilibrium** where the forward and backward rates perfectly balance each other.

### The Equilibrium Constant

To find out exactly where these concentrations settle, we need to look at the equilibrium condition. At equilibrium, the rate of the forward reaction equals the rate of the backward reaction:

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

By rearranging this, we get the equilibrium constant $K_c$:

$$K_c = \frac{k_f}{k_b} = \frac{[	ext{P}]_{eq}}{[	ext{R}]_{eq}}$$

The problem gives us a crucial piece of information: the backward rate constant is four times the forward rate constant ($k_b = 4k_f$). Let's substitute this into our ratio:

$$\frac{[	ext{P}]_{eq}}{[	ext{R}]_{eq}} = \frac{k_f}{4k_f} = \frac{1}{4}$$

This elegant little fraction tells us that at equilibrium, for every one molecule of $	ext{P}$, there are four molecules of $	ext{R}$. In other words, $[	ext{R}]_{eq} = 4[	ext{P}]_{eq}$.

### The Math of Conservation

Now, we need to find the exact fractional values of these concentrations relative to our starting amount, $[	ext{R}]_0$. Since matter cannot be created or destroyed, the total number of moles of $	ext{R}$ and $	ext{P}$ at any given moment must equal the initial moles of $	ext{R}$. This is the **Law of Conservation of Mass**.

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

We already know that $[	ext{R}]_{eq} = 4[	ext{P}]_{eq}$. Let's substitute that in:

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

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

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

And since $	ext{R}$ is four times $	ext{P}$:

$$[	ext{R}]_{eq} = 4 	imes 0.2 [	ext{R}]_0 = 0.8 [	ext{R}]_0$$

### Decoding the Graph

We have our final destinations! The concentration of $	ext{R}$ starts at $1.0$ (or $100\%$) and exponentially decays until it levels off at an asymptote of $0.8$. Conversely, the concentration of $	ext{P}$ starts at $0$ and exponentially grows until it levels off at an asymptote of $0.2$.

When we look at the given options, we are searching for the graph that perfectly mirrors this mathematical reality. Option (C) shows the curve for $[	ext{R}]/[	ext{R}]_0$ settling at the $0.8$ mark, and the curve for $[	ext{P}]/[	ext{R}]_0$ settling at the $0.2$ mark. It is a flawless visual representation of the kinetics we just calculated.

Select Answer:

Visualized Solution

$R ⇌ P$

$K_{c} = \frac{k _{f}}{k _{b}}$

$\frac{[ P ] _{e q}}{[ R ] _{e q}} = \frac{1}{4}$

$[R]_{e q} = 4 [P]_{e q}$

$[R]_{e q} + [P]_{e q} = [R]_{0}$

$[P]_{e q} = 0.2 [R]_{0}$

$[R]_{e q} = 0.8 [R]_{0}$

$\frac{[ R ] _{e q}}{[ R ] _{0}} = 0.8$

$t_{1/2} \propto \frac{1}{k _{f} + k _{b}}$

The Sigma Insight: Rate of Chemical Reaction

The Beauty of Reversible Reactions

The Equilibrium Constant

The Math of Conservation

Decoding the Graph

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R⇌P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kc​=kb​kf​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [R]eq​[P]eq​​=41​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [R]eq​=4[P]eq​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [R]eq​+[P]eq​=[R]0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [P]eq​=0.2[R]0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [R]eq​=0.8[R]0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [R]0​[R]eq​​=0.8

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1/2​∝kf​+kb​1​