For the reaction 2A + 3B + 3/2 C - 3P, which statement is correct?
(a) dn_A/dt = 2/3 dn_B/dt = 3/4

Q: Why do we divide by the stoichiometric coefficients?

We divide by the coefficients to get a single, unique 'Rate of Reaction' that is independent of which species we are measuring. Since 2 moles of A disappear for every 3 moles of B, A disappears slower than B. Dividing by their coefficients normalizes this.

Q: Why are there negative signs in the rate expression?

The change in moles ($dn$) for a reactant is negative because it is being consumed. We add a negative sign in the formula so that the overall 'Rate' is a positive value.

Q: Why did we drop the negative signs in the final step?

The options are comparing the magnitudes of the rates of disappearance. Since all reactants are disappearing, their individual $dn/dt$ values are all negative. Equating them directly without the negative signs is mathematically equivalent to equating their absolute values.

Q: What if the question asked for the rate of appearance of P?

You would simply include the product term in your equality: $\dots = +\frac{1}{3}\frac{dn_P}{dt}$. Then you could relate it to any of the reactants.

Q: Can we use concentrations instead of moles?

Yes! For a constant volume system, the rate of change of concentration $\frac{d[A]}{dt}$ is directly proportional to the rate of change of moles $\frac{dn_A}{dt}$. The stoichiometric relationships remain exactly the same.

Question

For the reaction $2A + 3B + \frac{3}{2}C \rightarrow 3P$, which statement is correct?

Options:

(A) $\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} = \frac{3}{4}\frac{dn_C}{dt}$

(B) $\frac{dn_A}{dt} = \frac{3}{2}\frac{dn_B}{dt} = \frac{3}{4}\frac{dn_C}{dt}$

(C) $\frac{dn_A}{dt} = \frac{dn_B}{dt} = \frac{dn_C}{dt}$

(D) $\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} = \frac{4}{3}\frac{dn_C}{dt}$

Sigma Percentile · Accepted Answer

The study of chemical kinetics begins with understanding how to express the rate of a reaction. It's not just about how fast a reactant disappears or a product appears; it's about how these individual rates relate to the overall stoichiometry of the reaction. Let's dive into a classic problem that tests exactly this concept.

### Analyzing the Setup

We are given the following chemical reaction:
$$2A + 3B + \frac{3}{2}C \longrightarrow 3P$$

Our goal is to find the correct mathematical relationship between the rates of disappearance of the reactants $A$, $B$, and $C$. The options are given in terms of the rate of change of moles, $\frac{dn}{dt}$.

### The Master Equation

To solve this, we need to recall the fundamental definition of the rate of a reaction. For any general reaction of the form $aA + bB \longrightarrow cC + dD$, the overall rate of the reaction is given by dividing the rate of change of each species by its stoichiometric coefficient:
$$	ext{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$$

The negative sign is used for reactants because their concentration (or number of moles) decreases over time, making the derivative negative. The negative sign ensures the overall rate is a positive quantity.

### Setting Up the Specifics

Let's apply this master equation to our specific reaction. We will write the rate in terms of the number of moles ($n$) instead of concentration, as requested by the options.
$$	ext{Rate} = -\frac{1}{2}\frac{dn_A}{dt} = -\frac{1}{3}\frac{dn_B}{dt} = -\frac{1}{3/2}\frac{dn_C}{dt}$$

Now, let's simplify the term for $C$. Dividing by a fraction is the same as multiplying by its reciprocal. So, $\frac{1}{3/2}$ becomes $\frac{2}{3}$. Also, since the options are comparing the magnitudes of these rates (all positive terms), we can drop the negative signs for the sake of comparison:
$$\frac{1}{2}\frac{dn_A}{dt} = \frac{1}{3}\frac{dn_B}{dt} = \frac{2}{3}\frac{dn_C}{dt}$$

### Final Calculation

We are almost there! If we look at the options, they all have $\frac{dn_A}{dt}$ isolated on the left side with a coefficient of $1$. To achieve this, we need to multiply our entire equation by $2$:
$$2 	imes \left( \frac{1}{2}\frac{dn_A}{dt} ight) = 2 	imes \left( \frac{1}{3}\frac{dn_B}{dt} ight) = 2 	imes \left( \frac{2}{3}\frac{dn_C}{dt} ight)$$

This simplifies beautifully to:
**$$\frac{dn_A}{dt} = \frac{2}{3}\frac{dn_B}{dt} = \frac{4}{3}\frac{dn_C}{dt}$$**

This perfectly matches option (d). The beauty of this method is that it's foolproof. As long as you carefully divide by the stoichiometric coefficients, you can relate the rate of any species in a reaction to any other species.

Animated Solution for Chemistry - Chemical Kinetics: For the reaction $2 A + 3 B + \frac{3}{2} C \to 3 P$ , which statement is correct?

Select Answer:

Visualized Solution

$The Chemical Reaction$

$General Rate Expression$

$Applying to Given Reaction$

$Simplifying the Expression$

$Isolating \frac{d n _{A}}{d t}$

$The Way Forward$

The Sigma Insight: Rate of Chemical Reaction

Analyzing the Setup

The Master Equation

Setting Up the Specifics

Final Calculation

Similar Questions

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

For the following reaction $2 X + Y k P$ the rate of reaction is $\frac{d [ P ]}{d t} = k [X]$ . Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are) (Use: $ln 2 = 0.693$ )

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

The rate law for a reaction between the substances $A$ and $B$ is given by $rate = k [A]^{n} [B]^{m}$ . On doubling the concentration of $A$ and halving the concentration of $B$ , the ratio of the new rate to the earlier rate of the reaction will be as

For the reaction, $H_{2} + I_{2} ⟶ 2 H I$ , the differential rate law is

For a first order reaction, $A ⟶ products$ the concentration of $A$ changes from $0.1 M$ to $0.025 M$ in $40 min$ . The rate of reaction when the concentration of $A$ is $0.01 M$ is

Which of the following plots is(are) correct for the given reaction? ( $[P]_{0}$ is the initial concentration of P)

For a first-order reaction R $⟶$ P at a given temperature, $k$ is the rate constant. For this reaction, at the given temperature, the concentrations of R and P at a time $t$ are [R] and [P], respectively. The correct graphical representation(s) for this reaction is(are)

The reaction $2 A + B_{2} \to 2 A B$ is an elementary reaction. For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ........... . (Round off to the nearest integer).

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 the reaction 2A+3B+23​C→3P, which statement is correct?

Select Answer:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying to Given Reaction

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

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

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