Question:
For the equilibrium, $X_2(g) + O_2(g) \rightleftharpoons 2XO(g)$, the equilibrium concentrations of $X_2(g)$ and $O_2(g)$ are $4 \times 10^{-3}M$ and $8 \times 10^{-3}M$ respectively. What is the equilibrium concentration of $XO(g)$? ($K_c = 0.5$)
For the equilibrium, $X_2(g) + O_2(g) \rightleftharpoons 2XO(g)$, the equilibrium concentrations of $X_2(g)$ and $O_2(g)$ are $4 \times 10^{-3}M$ and $8 \times 10^{-3}M$ respectively. What is the equilibrium concentration of $XO(g)$? ($K_c = 0.5$)
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Always square root when concentration is squared in $K_c$.
Updated On: Apr 24, 2026
- $4 \times 10^{-3}M$
- $6 \times 10^{-3}M$
- $5 \times 10^{-3}M$
- $2 \times 10^{-3}M$
- $8 \times 10^{-3}M$
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The Correct Option is A
Solution and Explanation
Concept:
• Equilibrium constant: \[ K_c = \frac{[XO]^2}{[X_2][O_2]} \]
Step 1: Substitute values
\[ 0.5 = \frac{[XO]^2}{(4\times10^{-3})(8\times10^{-3})} \]
Step 2: Solve
\[ [XO]^2 = 0.5 \times 32\times10^{-6} = 16\times10^{-6} \] \[ [XO] = 4\times10^{-3}M \] Final Conclusion:
Option (A)
• Equilibrium constant: \[ K_c = \frac{[XO]^2}{[X_2][O_2]} \]
Step 1: Substitute values
\[ 0.5 = \frac{[XO]^2}{(4\times10^{-3})(8\times10^{-3})} \]
Step 2: Solve
\[ [XO]^2 = 0.5 \times 32\times10^{-6} = 16\times10^{-6} \] \[ [XO] = 4\times10^{-3}M \] Final Conclusion:
Option (A)
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