Question:
If the rate of disappearance of \(\mathrm{N_2O_5}\) in the following reaction is \(1.2\times10^{-5}\ \mathrm{mol\ L^{-1}\ s^{-1}}\), the rate of production of \(\mathrm{NO_2}\) in \(\mathrm{mol\ L^{-1}\ s^{-1}}\) is
\[
2\mathrm{N_2O_5}(g)\rightarrow 4\mathrm{NO_2}(g)+\mathrm{O_2}(g)
\]
If the rate of disappearance of \(\mathrm{N_2O_5}\) in the following reaction is \(1.2\times10^{-5}\ \mathrm{mol\ L^{-1}\ s^{-1}}\), the rate of production of \(\mathrm{NO_2}\) in \(\mathrm{mol\ L^{-1}\ s^{-1}}\) is
\[ 2\mathrm{N_2O_5}(g)\rightarrow 4\mathrm{NO_2}(g)+\mathrm{O_2}(g) \]
\[ 2\mathrm{N_2O_5}(g)\rightarrow 4\mathrm{NO_2}(g)+\mathrm{O_2}(g) \]
Show Hint
For a balanced reaction, rates of disappearance and formation are related by stoichiometric coefficients. If
\[
aA\rightarrow bB,
\]
then
\[
\frac{d[B]}{dt}=\frac{b}{a}\left(-\frac{d[A]}{dt}\right).
\]
Updated On: Jun 26, 2026
- \(1.2\times10^{-5}\)
- \(3.6\times10^{-5}\)
- \(2.4\times10^{-5}\)
- \(4.8\times10^{-5}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Write the balanced chemical equation.
The given reaction is \[ 2\mathrm{N_2O_5}(g)\rightarrow 4\mathrm{NO_2}(g)+\mathrm{O_2}(g) \]
Step 2: Write the rate relation.
For the reaction, \[ 2\mathrm{N_2O_5}\rightarrow 4\mathrm{NO_2}+\mathrm{O_2} \] the rate relation is \[ -\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt} = \frac{1}{4}\frac{d[\mathrm{NO_2}]}{dt} \]
Step 3: Substitute the given rate of disappearance of \(\mathrm{N_2O_5}\).
Given, \[ -\frac{d[\mathrm{N_2O_5}]}{dt} = 1.2\times10^{-5}\ \mathrm{mol\ L^{-1}\ s^{-1}} \] Using the stoichiometric ratio, \[ \frac{d[\mathrm{NO_2}]}{dt} = \frac{4}{2} \left( -\frac{d[\mathrm{N_2O_5}]}{dt} \right) \] \[ \frac{d[\mathrm{NO_2}]}{dt} = 2\times 1.2\times10^{-5} \] \[ \frac{d[\mathrm{NO_2}]}{dt} = 2.4\times10^{-5}\ \mathrm{mol\ L^{-1}\ s^{-1}} \]
Step 4: Final conclusion.
Therefore, the rate of production of \(\mathrm{NO_2}\) is \[ \boxed{2.4\times10^{-5}\ \mathrm{mol\ L^{-1}\ s^{-1}}} \] Hence, the correct option is \[ \boxed{(3)} \]
The given reaction is \[ 2\mathrm{N_2O_5}(g)\rightarrow 4\mathrm{NO_2}(g)+\mathrm{O_2}(g) \]
Step 2: Write the rate relation.
For the reaction, \[ 2\mathrm{N_2O_5}\rightarrow 4\mathrm{NO_2}+\mathrm{O_2} \] the rate relation is \[ -\frac{1}{2}\frac{d[\mathrm{N_2O_5}]}{dt} = \frac{1}{4}\frac{d[\mathrm{NO_2}]}{dt} \]
Step 3: Substitute the given rate of disappearance of \(\mathrm{N_2O_5}\).
Given, \[ -\frac{d[\mathrm{N_2O_5}]}{dt} = 1.2\times10^{-5}\ \mathrm{mol\ L^{-1}\ s^{-1}} \] Using the stoichiometric ratio, \[ \frac{d[\mathrm{NO_2}]}{dt} = \frac{4}{2} \left( -\frac{d[\mathrm{N_2O_5}]}{dt} \right) \] \[ \frac{d[\mathrm{NO_2}]}{dt} = 2\times 1.2\times10^{-5} \] \[ \frac{d[\mathrm{NO_2}]}{dt} = 2.4\times10^{-5}\ \mathrm{mol\ L^{-1}\ s^{-1}} \]
Step 4: Final conclusion.
Therefore, the rate of production of \(\mathrm{NO_2}\) is \[ \boxed{2.4\times10^{-5}\ \mathrm{mol\ L^{-1}\ s^{-1}}} \] Hence, the correct option is \[ \boxed{(3)} \]
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