OnlineCBT
Chemistry · EQUILIBRIUM
हिन्दीAt \( 800\text{ K} \), equilibrium concentrations in a sealed vessel are \( [\text{N}_2] = 3.0 \times 10^{-3}\text{ M} \), \( [\text{O}_2] = 4.2 \times 10^{-3}\text{ M} \), and \( [\text{NO}] = 2.8 \times 10^{-3}\text{ M} \) for the reaction \( \text{N}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{NO}(\text{g}) \). If the reaction is written as \( 2\text{NO}(\text{g}) \rightleftharpoons \text{N}_2(\text{g}) + \text{O}_2(\text{g}) \), what is the numerical value of its equilibrium constant \( K_c' \)?
Correct answer · Option 2
1.608
Explanation
For forward reaction, \( K_c = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})} = 0.622 \). For the reverse reaction, \( K_c' = \frac{1}{K_c} = \frac{1}{0.622} \approx 1.608 \).
