Question

Consider a gas phase reaction which occurs in a closed vessel. \[ 2A (g) \rightarrow 4B (c) + C (g) \] The concentration of B is found to be increased by \( 5 \times 10^{-3} \) mol L\(^{-1}\) in 10 seconds. The rate of disappearance of A (in mol L\(^{-1}\) s\(^{-1}\)) is

• A. \( 4.75 \times 10^{-4} \)
• B. \( 7.5 \times 10^{-4} \)
• C. \( 1.25 \times 10^{-4} \)
• D. \( 2.5 \times 10^{-4} \)

Hint:

To relate the rate of disappearance of reactants and the rate of formation of products, use stoichiometry based on the balanced equation.

Answer 1

The Correct Option is D

To find the rate of disappearance of \( A \), we need to relate the changes in concentration of \( B \) to \( A \) using the stoichiometry of the reaction:

The reaction is:
\[ 2A (g) \rightarrow 4B (c) + C (g) \]

Given:

• The concentration of \( B \) increases by \( 5 \times 10^{-3} \) mol L\(^{-1}\) in 10 seconds.

Calculation:

1. Rate of formation of \( B \):
   \[ \text{Rate of } B = \frac{\Delta[B]}{\Delta t} = \frac{5 \times 10^{-3} \, \text{mol L}^{-1}}{10 \, \text{s}} = 5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]
2. From the balanced equation, 4 moles of \( B \) are produced for every 2 moles of \( A \) consumed. Therefore, the rate of disappearance of \( A \) is half the rate of formation of \( B \):
   \[ \text{Rate of disappearance of } A = \frac{1}{2} \times \text{Rate of } B = \frac{1}{2} \times 5 \times 10^{-4} = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

Conclusion:

The rate of disappearance of \( A \) is \( 2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}