Question:
Equal moles of NO$_2$(g) and NO(g) are to be placed in a container to produce N$_2$O according to the reaction,
\[
\text{NO}_2 + \text{NO} \rightleftharpoons \text{N}_2\text{O} + \text{O}_2, \quad K_c = 0.914
\]
How many moles of NO$_2$ and NO be placed in the 5.0 L container to have an equilibrium concentration of N$_2$O to be 0.05 M?
Equal moles of NO$_2$(g) and NO(g) are to be placed in a container to produce N$_2$O according to the reaction,
\[
\text{NO}_2 + \text{NO} \rightleftharpoons \text{N}_2\text{O} + \text{O}_2, \quad K_c = 0.914
\]
How many moles of NO$_2$ and NO be placed in the 5.0 L container to have an equilibrium concentration of N$_2$O to be 0.05 M?
Show Hint
Convert concentration to moles before applying $K_c$.
Updated On: Apr 23, 2026
- 0.5115
- 0.1023
- 0.0526
- 0.2046
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
Use equilibrium expression:
\[
K_c = \frac{[\text{N}_2\text{O}][\text{O}_2]}{[\text{NO}_2][\text{NO}]}
\]
Step 1: {Let initial moles of NO$_2$ and NO = $x$ each in 5 L.}
Step 2: {Given [N$_2$O] = 0.05 M $\Rightarrow$ moles = $0.05 \times 5 = 0.25$.}
Step 3: {From stoichiometry, NO$_2$ and NO consumed = 0.25 moles each.}
\[ [\text{NO}_2] = [\text{NO}] = \frac{x - 0.25}{5}, \quad [\text{O}_2] = 0.05 \]
Step 4: Substitute into $K_c$:
\[ 0.914 = \frac{0.05 \times 0.05}{\left(\frac{x - 0.25}{5}\right)^2} \]
Step 5: Solve:
\[ 0.914 = \frac{0.0625}{(x - 0.25)^2} \Rightarrow (x - 0.25)^2 \approx 0.0684 \] \[ x \approx 0.5115 \] Conclusion:
Required moles = 0.5115
Step 1: {Let initial moles of NO$_2$ and NO = $x$ each in 5 L.}
Step 2: {Given [N$_2$O] = 0.05 M $\Rightarrow$ moles = $0.05 \times 5 = 0.25$.}
Step 3: {From stoichiometry, NO$_2$ and NO consumed = 0.25 moles each.}
\[ [\text{NO}_2] = [\text{NO}] = \frac{x - 0.25}{5}, \quad [\text{O}_2] = 0.05 \]
Step 4: Substitute into $K_c$:
\[ 0.914 = \frac{0.05 \times 0.05}{\left(\frac{x - 0.25}{5}\right)^2} \]
Step 5: Solve:
\[ 0.914 = \frac{0.0625}{(x - 0.25)^2} \Rightarrow (x - 0.25)^2 \approx 0.0684 \] \[ x \approx 0.5115 \] Conclusion:
Required moles = 0.5115
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