Question

The rate constant, \( k \), of a zero order reaction \( 2 \text{NH}_3 (g) \xrightarrow{P_{1130K}} \text{N}_2 (g) + 3 \text{H}_2 (g) \) is \( y \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \). The rate of formation of hydrogen (in mol L\(^{-1}\) s\(^{-1}\)) is

• A. \( y \times 10^{-4} \)
• B. \( 2y \times 10^{-4} \)
• C. \( 3y \times 10^{-4} \)
• D. \( \frac{y \times 10^{-4}}{3} \)

Hint:

For zero-order reactions, the rate is independent of the concentration of reactants, and the rate of formation of products is directly proportional to the rate constant.

Answer 1

The Correct Option is C

For a zero order reaction, the rate of reaction is independent of the concentration of reactants. The rate law for the reaction \( 2 \text{NH}_3 (g) \xrightarrow{P_{1130K}} \text{N}_2 (g) + 3 \text{H}_2 (g) \) can be written as: \[ \text{Rate} = k \times [\text{NH}_3]^0 = k \] Given that the rate constant \( k = y \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \), the rate of hydrogen formation is given by: \[ \text{Rate of hydrogen formation} = 3 \times \text{Rate of reaction} = 3 \times k = 3 \times y \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \] Thus, the rate of hydrogen formation is \( 3y \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \). The correct answer is option (3).

Answer 2

Step 1: Understand the reaction and rate law
The given reaction is:
\[ 2 \text{NH}_3 (g) \xrightarrow{P_{1130K}} \text{N}_2 (g) + 3 \text{H}_2 (g) \]
It is a zero order reaction with rate constant \( k = y \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \).

Step 2: Write the rate expression
For a zero order reaction, the rate of disappearance of ammonia is:
\[ \text{Rate} = k = y \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

Step 3: Relate rates of reactants and products
From the stoichiometry:
- 2 moles of NH\(_3\) produce 3 moles of H\(_2\)
Therefore, rate of formation of H\(_2\) is related to rate of disappearance of NH\(_3\) by:
\[ \text{Rate of } \mathrm{H}_2 = \frac{3}{2} \times \text{Rate of } \mathrm{NH}_3 \]

Step 4: Calculate rate of H\(_2\) formation
Given rate of disappearance of NH\(_3\) = \( k = y \times 10^{-4} \)
\[ \text{Rate of } \mathrm{H}_2 = \frac{3}{2} \times y \times 10^{-4} \]
But since rate is often expressed per mole of reaction (considering 2 moles NH\(_3\) react), the rate of formation of H\(_2\) is:
\[ 3 \times y \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

Step 5: Conclusion
The rate of formation of hydrogen gas is \( 3y \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \).