Question

For the reaction, $N _{2}+3 H _{2} \rightarrow 2 NH _{3}$, If $\frac{ d \left[ NH _{3}\right]}{ dt }=2 \times 10^{-4}\, mol\, L ^{-1} s ^{-1}$, The value of $\frac{- d \left[ H _{2}\right]}{ dt }$ would be -

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

Answer 1

The Correct Option is A

To solve this question, we need to understand the stoichiometry of the given chemical reaction and apply it to calculate the rate of change of concentration for hydrogen gas, H_2.

Step-by-Step Solution:

The balanced chemical reaction is:

N_2 + 3H_2 \rightarrow 2NH_3

Given:

• The rate of formation of ammonia, \frac{d[NH_3]}{dt} = 2 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}.

According to the stoichiometry of the reaction:

• 1 mole of N_2 reacts with 3 moles of H_2 to produce 2 moles of NH_3.

The rate of change for the reactants and products can be expressed using their stoichiometric coefficients:

• \frac{-d[N_2]}{dt} = \frac{1}{2} \times \frac{d[NH_3]}{dt}
• \frac{-d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt}

Substituting the given rate of change of ammonia:

\frac{-d[H_2]}{dt} = \frac{3}{2} \times 2 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}

Calculate the result:

\frac{-d[H_2]}{dt} = 3 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}

Therefore, the value of \frac{-d[H_2]}{dt} is:

• 3 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}.

Conclusion:

The correct answer is $3 \times 10^{-4}\,mol \, L^{-1} \, s^{-1}$.