Question

For the reaction \(\mathrm{P}+\mathrm{Q} \rightarrow 2 \mathrm{R}+\mathrm{S}\). Which of the following statement is/are correct? a. Rate of disappearance of \(\mathrm{P}=\) rate of appearance of \(\mathrm{S}\) b. Rate of disappearance of \(\mathrm{P}=\) rate of disappearance of \(\mathrm{Q}\) c. Rate of disappearance of \(\mathrm{Q}=2 \times\) rate of appearance of \(\mathrm{R}\) d. Rate of disappearance of \(\mathrm{Q}=1 / 2 \times\) rate of appearance of \(\mathrm{R}\)

Short answer

Statements a, b, and c are correct; statement d is incorrect.

Step-by-step solution

Understanding the Reaction

The chemical reaction is represented as: \[ \mathrm{P} + \mathrm{Q} \rightarrow 2 \mathrm{R} + \mathrm{S} \]. We need to compare the rates of disappearance and appearance of the reactants and products.

Analyzing Stoichiometry

From the balanced chemical equation, one mole of \( \mathrm{P} \) and one mole of \( \mathrm{Q} \) react to form two moles of \( \mathrm{R} \) and one mole of \( \mathrm{S} \). We'll use the stoichiometric coefficients to compare reaction rates.

Comparing Rates for Statement a

According to stoichiometry, for every mole of \( \mathrm{P} \) disappearing, one mole of \( \mathrm{S} \) forms, so the rate of disappearance of \( \mathrm{P} \) equals the rate of appearance of \( \mathrm{S} \). Statement a is correct.

Comparing Rates for Statement b

Here, \( \mathrm{P} \) and \( \mathrm{Q} \) have a 1:1 stoichiometry. Thus, their rates of disappearance are also equal. Statement b is correct.

Comparing Rates for Statements c and d

For statement c, the rate of disappearance of \( \mathrm{Q} \) would equal the rate of appearance of \( \mathrm{R} \) times the stoichiometric ratio, which is 2:1. Therefore, statement c is correct. But statement d incorrectly states the inverse, so it is incorrect.

Key concepts

Rate of Disappearance

The rate of disappearance refers to how quickly a reactant is used up during a chemical reaction. For the reaction \( \mathrm{P} + \mathrm{Q} \rightarrow 2 \mathrm{R} + \mathrm{S} \), the rate at which the reactants \( \mathrm{P} \) and \( \mathrm{Q} \) disappear can be expressed based on their consumption over time. If we track the concentration of \( \mathrm{P} \) and \( \mathrm{Q} \):

• The rate of disappearance of \( \mathrm{P} \) is given by \( \frac{-d[\mathrm{P}]}{dt} \).
• The rate of disappearance of \( \mathrm{Q} \) is given by \( \frac{-d[\mathrm{Q}]}{dt} \).

The negative sign indicates that their concentration decreases as the reaction proceeds. In our example, statements b and c correctly use this concept to assert equality or proportionality in rate. Since \( \mathrm{P} \) and \( \mathrm{Q} \) are consumed at the same rates due to their 1:1 stoichiometric ratio, their rate of disappearance is equal.

Rate of Appearance

The rate of appearance refers to how quickly a product forms in a chemical reaction. It is typically expressed as the change in concentration of a product over time. For the reaction \( \mathrm{P} + \mathrm{Q} \rightarrow 2 \mathrm{R} + \mathrm{S} \), the rates at which the products \( \mathrm{R} \) and \( \mathrm{S} \) appear can be expressed as:

• The rate of appearance of \( \mathrm{R} \) is \( \frac{d[\mathrm{R}]}{dt} \).
• The rate of appearance of \( \mathrm{S} \) is \( \frac{d[\mathrm{S}]}{dt} \).

For this reaction, \( \mathrm{S} \) and \( \mathrm{P} \) appear and disappear at the same rate because one mole of \( \mathrm{P} \) produces one mole of \( \mathrm{S} \). This is why statement a holds true. However, when it comes to \( \mathrm{R} \), two moles are produced for each mole of \( \mathrm{Q} \) that disappears, making its rate of appearance twice that of the rate of disappearance of \( \mathrm{Q} \). Statement c accurately reflects this principle.

Stoichiometry

Stoichiometry in a chemical reaction provides the relationship between quantities of reactants and products. It is governed by the coefficients found in the balanced chemical equation. In the reaction \( \mathrm{P} + \mathrm{Q} \rightarrow 2 \mathrm{R} + \mathrm{S} \), stoichiometry tells us:

• 1 mole of \( \mathrm{P} \) reacts with 1 mole of \( \mathrm{Q} \).
• 2 moles of \( \mathrm{R} \) and 1 mole of \( \mathrm{S} \) are produced.

This helps determine the rate relationships. Each coefficient acts as a multiplier for calculating the rates. For instance, given that 2 moles of \( \mathrm{R} \) form per mole of \( \mathrm{Q} \), we see how \( \mathrm{R} \)'s production rate is double the consumption rate of \( \mathrm{Q} \). Stoichiometry not only balances chemical equations but also ensures accurate rate comparisons which is critical in analyzing any chemical reaction's kinetics.