The equilibrium constant for the reaction: (g) + (g) β 2(g) is K1 and the equilibrium constant for t β Chemical Equilibrium Chemistry Question
Question
The equilibrium constant for the reaction: $N_2$(g) + $O_2$(g) β 2$NO$(g) is K1 and the equilibrium constant for the reaction: $NO$(g) β 1/2 $N_2$(g) + 1/2 $O_2$(g) is K2, both at the same temperature. K1 and K2 are related as
π‘ Solution & Explanation
Step 1 - Express the equilibrium constants of both reactions We are given two reversible chemical reactions at the same temperature: 1. **First Reaction:** \[\ce{N2(g) + O2(g) <=> 2NO(g)}\] According to the law of chemical equilibrium, the equilibrium constant ($K_1$) is expressed as: \[K_1 = \frac{[\ce{NO}]^2}{[\ce{N2}][\ce{O2}]}\] 2. **Second Reaction:** \[\ce{NO(g) <=> 1/2 N2(g) + 1/2 O2(g)}\] The equilibrium constant ($K_2$) for this reaction is: \[K_2 = \frac{[\ce{N2}]^{1/2} [\ce{O2}]^{1/2}}{[\ce{NO}]}\] Step 2 - Relate the two reactions using rules of multiple equilibria By comparing the two chemical equations, we can see that the second reaction is obtained by manipulating the first reaction in two steps: 1. **Reversing the first reaction:** \[\ce{2NO(g) <=> N2(g) + O2(g)}\] When a reaction is reversed, its equilibrium constant is inverted: \[K_{\text{rev}} = \frac{1}{K_1}\] 2. **Multiplying the stoichiometry of the reversed reaction by a factor of $\frac{1}{2}$:** \[\ce{NO(g) <=> 1/2 N2(g) + 1/2 O2(g)}\] When the stoichiometric coefficients of a reaction are multiplied by a factor $n$, the new equilibrium constant is raised to the power of $n$. Here, $n = \frac{1}{2}$: \[K_2 = \left(K_{\text{rev}}\right)^{1/2} = \left(\frac{1}{K_1}\right)^{1/2} = \frac{1}{\sqrt{K_1}}\] Step 3 - Perform algebraic manipulation to match the options We now have the relationship: \[K_2 = \frac{1}{\sqrt{K_1}}\] To solve for $K_1$, we first square both sides of the equation: \[K_2^2 = \frac{1}{K_1}\] Taking the reciprocal of both sides: \[K_1 = \frac{1}{K_2^2} = \left(\frac{1}{K_2}\right)^2\] Step 4 - Evaluate the options systematically * **Option (A) $K_1 = (1 / K_2)^2$**: Correct. As derived, this matches our relationship $K_1 = K_2^{-2}$. * **Option (B) $K_1 = K_2^2$**: Incorrect. This would be true if the second reaction was half of the first reaction but not reversed (which would mean $K_2 = \sqrt{K_1}$ or $K_1 = K_2^2$). * **Option (C) $K_2 = (1 / K_1)^2$**: Incorrect. This would represent a reaction that was reversed and doubled in stoichiometry, yielding $K_2 = K_1^{-2}$. * **Option (D) $K_2 = K_1^2$**: Incorrect. This represents a reaction whose stoichiometry was doubled without being reversed, yielding $K_2 = K_1^2$. \[\boxed{\text{A}}\]