Question

At equilibrium the concentrations of ${\mathrm{N}}_2=3.0\times {10}^{-3}\mathrm{M},{\mathrm{O}}_2=4.2\times {10}^{-3}\mathrm{M}$ and $\mathrm{NO}=2.8\times {10}^{-3}\mathrm{M}$ in a sealed vessel at $800\mathrm{K}$. What will be ${\mathrm{K}}_{\mathrm{C}}$ for the reaction ${\mathrm{N}}_{2(\mathrm{g})}+{\mathrm{O}}_{2(\mathrm{g})}\leftrightharpoons 2{\mathrm{NO}}_{(\mathrm{g})}?$

• (A) $0.622$
• (B) $0.722$
• (C) $0.822$
• (D) $0.922$

Answer 1

The Correct Option is A

Explanation:
We are given,${\mathrm{N}}_2=3.0\times {10}^{-3}\mathrm{M}$${\mathrm{O}}_2=4.2\times {10}^{-3}\mathrm{M}$$\mathrm{NO}=2.8\times {10}^{-3}\mathrm{M}$Applying the law of chemical equilibrium, we will get -${\mathrm{K}}_{\mathrm{C}}=\frac{[\mathrm{NO}{]}^2}{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]}\cdots (1)$Adding the values in equation 1 ,${\mathrm{K}}_{\mathrm{C}}=\frac{[2.8\times {10}^{-3}\mathrm{M}]}{[3.0\times {10}^{-3}\mathrm{M}][4.2\times {10}^{-3}\mathrm{M}]}=0.622$The value of ${\mathrm{K}}_{\mathrm{C}}$ is $0.622$Hence, the correct option is (A).