For a reaction 1/2 A - 2B, rate of disappearance of 'A' is related to the rate of appearance of 'B'

Q: Why is there a negative sign for A?

Reactants disappear over time, so their change in concentration $\frac{d[A]}{dt}$ is negative. The negative sign makes the overall rate positive.

Q: Why do we divide by the stoichiometric coefficient?

To normalize the rate so that the "Rate of Reaction" is a single, unique value regardless of which species we are tracking.

Q: Can a stoichiometric coefficient be a fraction?

Yes, in balanced equations, fractions are perfectly valid and represent molar ratios.

Q: What does "rate of disappearance" exactly mean?

It is simply the magnitude of the rate at which the reactant is consumed, represented by $-\frac{d[A]}{dt}$ without the stoichiometric coefficient.

Q: I got option (d) by multiplying instead of dividing. Why is that wrong?

The rate of reaction is defined per mole of the reaction as written. If 2 moles of B are produced for every 0.5 moles of A consumed, B appears 4 times faster than A disappears. Thus, the rate of A is 1/4th the rate of B.

Question

For a reaction $\frac{1}{2}A \longrightarrow 2B$, rate of disappearance of 'A' is related to the rate of appearance of 'B' by the expression

Options:

(A) $-\frac{d[A]}{dt} = \frac{1}{2}\frac{d[B]}{dt}$

(B) $-\frac{d[A]}{dt} = \frac{1}{4}\frac{d[B]}{dt}$

(C) $-\frac{d[A]}{dt} = \frac{d[B]}{dt}$

(D) $-\frac{d[A]}{dt} = 4\frac{d[B]}{dt}$

Sigma Percentile · Accepted Answer

### The Chemical Setup

Imagine a bustling factory where raw materials are being consumed to produce finished goods. In our chemical reaction, the reactant $A$ is the raw material, and $B$ is the product.

The balanced equation is given as:

$$ \frac{1}{2}A \longrightarrow 2B $$

This tells us that for every half mole of $A$ that disappears, two moles of $B$ are formed. The stoichiometry is the blueprint of this factory, dictating the exact ratio of consumption to production.

### The Golden Rule of Rates

When we talk about the **Rate of Reaction**, we want a single, universal number that describes how fast the overall process is happening, regardless of whether we are looking at $A$ or $B$.

To achieve this, we must normalize the individual rates. The golden rule is to divide the rate of change of each species by its respective stoichiometric coefficient.

For a general reaction $aA \longrightarrow bB$, the universal rate is expressed as:

$$ -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} $$

The negative sign for $A$ is crucial. Since $A$ is a reactant, its concentration decreases over time, making $\frac{d[A]}{dt}$ a negative value. The extra minus sign ensures the overall rate remains a positive quantity.

### The Mathematical Execution

Now, let's apply this golden rule to our specific reaction. We identify our stoichiometric coefficients as $a = \frac{1}{2}$ and $b = 2$.

Substituting these into our rate law gives:

$$ -\frac{1}{\frac{1}{2}}\frac{d[A]}{dt} = \frac{1}{2}\frac{d[B]}{dt} $$

Let's simplify the complex fraction on the left side. Dividing by a fraction is the same as multiplying by its reciprocal.

$$ -2\frac{d[A]}{dt} = \frac{1}{2}\frac{d[B]}{dt} $$

### Isolating the Target

The question specifically asks for the **rate of disappearance of A**. This physical quantity is represented purely by $-\frac{d[A]}{dt}$, without any coefficients attached.

To isolate this term, we simply divide both sides of our equation by $2$.

$$ -\frac{d[A]}{dt} = \frac{1}{4}\frac{d[B]}{dt} $$

And there we have it! The rate at which $A$ disappears is exactly one-fourth the rate at which $B$ appears. This perfectly matches option (b).

Animated Solution for Chemistry - Chemical Kinetics: For a reaction $\frac{1}{2} A ⟶ 2 B$ , rate of disappearance of 'A' is related to the rate of appearance of 'B' by the expression

Select Answer:

Visualized Solution

Reaction Setup

General Rate Law

Substitute Coefficients

Simplify Expression

Isolate Rate of Disappearance

Key Takeaway

The Sigma Insight: Rate of Chemical Reaction

The Chemical Setup

The Golden Rule of Rates

The Mathematical Execution

Isolating the Target

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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 a reaction 21​A⟶2B, rate of disappearance of 'A' is related to the rate of appearance of 'B' by the expression