Question

Ammonia and oxygen react at high temperature as:
$4\text{NH}_{3(\text{g})} + 5\text{O}_{2(\text{g})} \longrightarrow 4\text{NO}_{(\text{g})} + 6\text{H}_2\text{O}_{(\text{g})}$
If rate of formation of $\text{NO}_{(\text{g})}$ is $3.6 \times 10^{-3}\ \text{mol\ L}^{-1}\text{s}^{-1}$ then rate of disappearance of ammonia is

• A. $7.2 \times 10^{-3}\ \text{mol\ L}^{-1}\text{s}^{-1}$
• B. $1.2 \times 10^{-3}\ \text{mol\ L}^{-1}\text{s}^{-1}$
• C. $2.4 \times 10^{-3}\ \text{mol\ L}^{-1}\text{s}^{-1}$
• D. $3.6 \times 10^{-3}\ \text{mol\ L}^{-1}\text{s}^{-1}$

Hint:

Whenever two species in a balanced chemical equation share the exact same stoichiometric coefficient (here, both $\text{NH}_3$ and NO have a coefficient of 4), their absolute chemical rates of consumption and formation will always be identical!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question gives a balanced chemical equation for the oxidation of ammonia and provides the rate of formation of nitric oxide (NO). We need to find the corresponding rate of disappearance of ammonia ($\text{NH}_3$).

Step 2: Key Formula or Approach:
From the principles of chemical kinetics, the overall rate of a reaction is related to the individual rates of consumption of reactants and formation of products divided by their respective stoichiometric coefficients:
$$\text{Rate} = -\frac{1}{4}\frac{\text{d}[\text{NH}_3]}{\text{d}t} = -\frac{1}{5}\frac{\text{d}[\text{O}_2]}{\text{d}t} = +\frac{1}{4}\frac{\text{d}[\text{NO}]}{\text{d}t} = +\frac{1}{6}\frac{\text{d}[\text{H}_2\text{O}]}{\text{d}t}$$

Step 3: Detailed Explanation:
Let's isolate the mathematical relationship connecting ammonia and nitric oxide from the main rate expression:
$$-\frac{1}{4}\frac{\text{d}[\text{NH}_3]}{\text{d}t} = +\frac{1}{4}\frac{\text{d}[\text{NO}]}{\text{d}t}$$ Multiply both sides of the equation by 4:
$$-\frac{\text{d}[\text{NH}_3]}{\text{d}t} = +\frac{\text{d}[\text{NO}]}{\text{d}t}$$ This equality tells us that the rate of disappearance of ammonia is perfectly equal to the rate of formation of nitric oxide.
Given that the rate of formation of NO is $3.6 \times 10^{-3}\ \text{mol\ L}^{-1}\text{s}^{-1}$, the rate of disappearance of $\text{NH}_3$ must also be exactly $3.6 \times 10^{-3}\ \text{mol\ L}^{-1}\text{s}^{-1}$. This matches option (D).

Step 4: Final Answer:
The rate of disappearance of ammonia is $3.6 \times 10^{-3}\ \text{mol\ L}^{-1}\text{s}^{-1}$, which corresponds to option (D).