Question

For the reaction \( N_2 + 3H_2 \rightarrow 2NH_3 \), if \( \frac{\Delta [NH_3]}{\Delta t} = 2 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \), then the value of \( \frac{\Delta [H_2]}{\Delta t} \) would be:

• A. \( 1 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)
• B. \( 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)
• C. \( 4 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)
• D. \( 6 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \)

Hint:

In rate law calculations, use the stoichiometric coefficients to determine the relationship between reactants and products.

Answer 1

The Correct Option is B

From the balanced chemical equation, the mole ratio of \( NH_3 \) to \( H_2 \) is \( 2:3 \). Therefore, the rate of change of \( H_2 \) is \( 3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \).
Final Answer: \[ \boxed{3 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}} \]