Question

For the reaction \(4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}\), if the rate of disappearance of \(\text{NH}_3\) is \(3.6 \times 10^{-3}\,\text{M s}^{-1}\), what is the rate of formation of water?

• A. \(4.0 \times 10^{4}\,\text{M s}^{-1}\)
• B. \(3.6 \times 10^{-3}\,\text{M s}^{-1}\)
• C. \(6.0 \times 10^{-4}\,\text{M s}^{-1}\)
• D. \(5.4 \times 10^{-3}\,\text{M s}^{-1}\)

Hint:

Always use stoichiometric coefficients to relate rates of reactants and products.

Answer 1

The Correct Option is D

Step 1: Write the stoichiometric rate relation.
For the reaction \[ 4\text{NH}_3 \rightarrow 6\text{H}_2\text{O} \] the rates are related as: \[ \frac{1}{4}\left(-\frac{d[\text{NH}_3]}{dt}\right) = \frac{1}{6}\left(\frac{d[\text{H}_2\text{O}]}{dt}\right) \] Step 2: Substitute the given rate.
\[ -\frac{d[\text{NH}_3]}{dt} = 3.6 \times 10^{-3}\,\text{M s}^{-1} \] Step 3: Calculate the rate of formation of water.
\[ \frac{d[\text{H}_2\text{O}]}{dt} = \frac{6}{4} \times 3.6 \times 10^{-3} = 5.4 \times 10^{-3}\,\text{M s}^{-1} \] Step 4: Conclusion.
Hence, the rate of formation of water is \(5.4 \times 10^{-3}\,\text{M s}^{-1}\).