Question

Nitric oxide reacts with H$_2$ according to reaction. 2NO$_{(\text{g})}$ + 2H$_{2(\text{g})}$ $\to$ N$_{2(\text{g})}$ + 2H$_2$O$_{(\text{g})}$.
Identify the correct relationship for consumption of reactant and formation of product.

• A. $\frac{1}{2} \frac{\text{d[NO]}}{\text{dt}} = - \frac{\text{d[H}_2]}{\text{dt}}$
• B. $\frac{\text{d[N}_2]}{\text{dt}} = - \frac{1}{2} \frac{\text{d[H}_2]}{\text{dt}}$
• C. $\frac{\text{d[H}_2\text{O]}}{\text{dt}} = \frac{\text{d[N}_2]}{\text{dt}}$
• D. $\frac{\text{d[H}_2\text{O]}}{\text{dt}} = \frac{1}{2} \frac{\text{d[N}_2]}{\text{dt}}$

Hint:

Reactant rates are negative (consumption); Product rates are positive (formation). Divide by coefficients.

Answer 1

The Correct Option is B

Step 1: Rate Expression
For reaction $aA + bB \to cC + dD$:
$\text{Rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$
Step 2: Apply to Reaction
$\text{Rate} = -\frac{1}{2} \frac{d[NO]}{dt} = -\frac{1}{2} \frac{d[H_2]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{2} \frac{d[H_2O]}{dt}$
Step 3: Verification
- Option (B) says $\frac{d[N_2]}{dt} = -\frac{1}{2} \frac{d[H_2]}{dt}$. This matches the derived expression.
Final Answer: (B)