For an elementary chemical reaction, A2 ⇌ 2A (forward rate constant k1, backward rate constant k-1),

Q: Why is the concentration of $A$ squared in the backward rate?

Because the problem specifies it's an *elementary* reaction. In elementary reactions, the reaction happens in a single step, so the stoichiometric coefficient directly becomes the power in the rate law. Since two $A$ atoms react, it's $[A]^2$.

Q: Why do we divide by the stoichiometric coefficient to find the rate of reaction?

To ensure the rate of reaction is a unique, unambiguous value regardless of which chemical species we are monitoring. Since $A$ is produced twice as fast as $A_2$ is consumed, dividing by 2 normalizes the rate.

Q: What exactly does "elementary reaction" mean here?

An elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step and with a single transition state. There are no hidden intermediate steps.

Q: Why is the net rate $r_f - r_b$?

The forward reaction ($r_f$) pushes the system towards the products, while the backward reaction ($r_b$) pushes it back towards the reactants. The net progress is the difference between these two opposing forces.

Q: Can we write the rate of reaction in terms of $A_2$?

Yes! Since the stoichiometric coefficient of $A_2$ is 1 and it's a reactant, the rate of reaction $r = -\frac{d[A_2]}{dt}$. This means $-\frac{d[A_2]}{dt} = k_1[A_2] - k_{-1}[A]^2$.

Question

For an elementary chemical reaction, $A_2 \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} 2A$, the expression for $\frac{d[A]}{dt}$ is

Options:

(A) $2k_1[A_2] - k_{-1}[A]^2$

(B) $k_1[A_2] - k_{-1}[A]^2$

(C) $2k_1[A_2] - 2k_{-1}[A]^2$

(D) $k_1[A_2] + k_{-1}[A]^2$

Sigma Percentile · Accepted Answer

The problem asks us to find the rate of appearance of product $A$ for the elementary reversible reaction $A_2 \rightleftharpoons 2A$. Let's break down the kinetics of this molecular dance step by step.

### The Reversible Dance of Molecules
Imagine a closed vessel where diatomic molecules of $A_2$ are constantly breaking apart into individual $A$ atoms, while simultaneously, pairs of $A$ atoms are colliding and recombining to form $A_2$. This is a dynamic, reversible process.

Because the problem states this is an **elementary reaction**, it means the reaction occurs exactly as written in a single step. For elementary reactions, the order of the reaction with respect to each reactant is simply its stoichiometric coefficient in the balanced equation.

### Formulating the Rates
Let's look at the forward and backward processes independently.

For the forward reaction, $A_2 \rightarrow 2A$, the rate depends only on the concentration of $A_2$. We can write the forward rate $r_f$ as:
$$r_f = k_1[A_2]$$

For the backward reaction, $2A \rightarrow A_2$, two atoms of $A$ must collide. Therefore, the rate depends on the concentration of $A$ squared. We can write the backward rate $r_b$ as:
$$r_b = k_{-1}[A]^2$$

The **net rate of reaction**, $r$, is the difference between the rate at which the forward reaction proceeds and the rate at which the backward reaction opposes it:
$$r = r_f - r_b = k_1[A_2] - k_{-1}[A]^2$$

### The Stoichiometric Connection
Now, we need to connect this overall rate $r$ to the specific rate of appearance of $A$, which is $\frac{d[A]}{dt}$.

According to the principles of chemical kinetics, the unique rate of a reaction is given by the rate of change of any species divided by its stoichiometric coefficient (with a positive sign for products and a negative sign for reactants).

For our product $A$, the stoichiometric coefficient is $2$. Therefore:
$$r = \frac{1}{2} \frac{d[A]}{dt}$$

### The Final Expression
To find the rate of appearance of $A$, we simply rearrange the equation to solve for $\frac{d[A]}{dt}$:
$$\frac{d[A]}{dt} = 2r$$

Substituting our expression for the net rate $r$:
$$\frac{d[A]}{dt} = 2 \left( k_1[A_2] - k_{-1}[A]^2 \right)$$

Expanding the bracket, we get our final answer:
$$\frac{d[A]}{dt} = 2k_1[A_2] - 2k_{-1}[A]^2$$

This elegant expression perfectly captures the dual nature of the reaction: $A$ is being produced at twice the forward rate and consumed at twice the backward rate!

Animated Solution for Chemistry - Chemical Kinetics: For an elementary chemical reaction, $A_{2} k_{- 1} ⇌ k_{1} 2 A$ , the expression for $\frac{d [ A ]}{d t}$ is

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Visualized Solution

$A_{2} ⇌ 2 A$

$Forward Rate$

$Backward Rate$

$Net Rate of Reaction$

$Rate and Stoichiometry$

$Rate of Appearance of A$

$Final Expression$

$Equilibrium Condition$

The Sigma Insight: Rate of Chemical Reaction

The Reversible Dance of Molecules Imagine a closed vessel where diatomic molecules of $A_{2}$ are constantly breaking apart into individual $A$ atoms, while simultaneously, pairs of $A$ atoms are colliding and recombining to form $A_{2}$

Formulating the Rates

The Stoichiometric Connection

The Final Expression

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Animated Solution for Chemistry - Chemical Kinetics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For an elementary chemical reaction, A2​k−1​⇌k1​​​2A, the expression for dtd[A]​ is

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​⇌2A

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Rate of Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate and Stoichiometry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate of Appearance of A

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

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