Question

For the reaction

\[ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})} \]

\(\mathrm{N}_{2(\mathrm{~g})}\) disappears at the rate of \(2 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\). Find the rate of \(\mathrm{NH}_3\) formation.

• A. \(4\times 10^{-4}\text{ }\mathrm{moldm}^{-3}\text{ }\mathrm{s}^{-1}\)
• B. \(1.5\times 10^{-4}\text{ }\mathrm{moldm}^{-3}\text{ }\mathrm{s}^{-1}\)
• C. \(2.0\times 10^{-4}\text{ }\mathrm{moldm}^{-3}\text{ }\mathrm{s}^{-1}\)
• D. \(2.5\times 10^{-4}\text{ }\mathrm{moldm}^{-3}\text{ }\mathrm{s}^{-1}\)

Hint:

Relate the rates of the reactants and products using the stoichiometric coefficients of the balanced equation.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For the reaction \(\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\), the rate is the same when each species rate is divided by its coefficient.
So \(-\dfrac{d[\text{N}_2]}{dt} = \dfrac{1}{2}\dfrac{d[\text{NH}_3]}{dt}\).

Step 2: Calculation
Given \(-\dfrac{d[\text{N}_2]}{dt} = 2\times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\).
Therefore \(\dfrac{d[\text{NH}_3]}{dt} = 2\times 2\times 10^{-4} = 4\times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\).

Step 3: Compare the options
One mole of \(\text{N}_2\) gives two moles of \(\text{NH}_3\), so the product forms twice as fast as nitrogen disappears. This gives \(4\times 10^{-4}\).

Final Answer:
The rate of formation of ammonia is \(4\times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). \[ \boxed{\text{(A) }4\times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}} \]