XeF6 + β XeOF4 + 2; K1. XeO4 + XeF6 β XeOF4 + XeO3; K2. The equilibrium constant for XeO4 + 2 β XeO3 β Chemical Equilibrium Chemistry Question
Question
XeF6 + $H_2O$ β XeOF4 + 2$HF$; K1. XeO4 + XeF6 β XeOF4 + XeO3$F_2$; K2. The equilibrium constant for XeO4 + 2$HF$ β XeO3$F_2$ + $H_2O$ is
π‘ Solution & Explanation
Step 1 - Express the equilibrium constants of the given reactions We are given two reversible gas-phase reactions with their respective equilibrium constants: Reaction (1): \[\ce{XeF6 + H2O <=> XeOF4 + 2HF} \quad \text{with equilibrium constant } K_1\] The equilibrium constant expression is: \[K_1 = \frac{[\ce{XeOF4}][\ce{HF}]^2}{[\ce{XeF6}][\ce{H2O}]}\] Reaction (2): \[\ce{XeO4 + XeF6 <=> XeOF4 + XeO3F2} \quad \text{with equilibrium constant } K_2\] The equilibrium constant expression is: \[K_2 = \frac{[\ce{XeOF4}][\ce{XeO3F2}]}{[\ce{XeO4}][\ce{XeF6}]}\] Step 2 - Identify the target reaction and its equilibrium constant expression The target reaction whose equilibrium constant (\(K_3\)) we need to find is: \[\ce{XeO4 + 2HF <=> XeO3F2 + H2O}\] The equilibrium constant expression for this target reaction is: \[K_3 = \frac{[\ce{XeO3F2}][\ce{H2O}]}{[\ce{XeO4}][\ce{HF}]^2}\] Step 3 - Manipulate the given equations to obtain the target equation 1. **Reverse Reaction (1):** Reversing a chemical reaction inverts its equilibrium constant. \[\ce{XeOF4 + 2HF <=> XeF6 + H2O} \quad \text{with equilibrium constant } K'_1 = \frac{1}{K_1}\] 2. **Add Reaction (2) to the reversed Reaction (1):** When we add two chemical equations, their corresponding equilibrium constants are multiplied. After adding: \[\begin{array}{rll} \ce{XeOF4 + 2HF} &\ce{<=> XeF6 + H2O} & \quad \left(K'_1 = \frac{1}{K_1}\right) \\ \ce{XeO4 + XeF6} &\ce{<=> XeOF4 + XeO3F2} & \quad (K_2) \\ \hline \end{array}\] Step 4 - Simplify the combined equation We cancel the common species appearing on both the reactant and product sides: - \(\ce{XeOF4}\) appears on both sides and cancels out. - \(\ce{XeF6}\) appears on both sides and cancels out. After cancellation, the net equation is: \[\ce{XeO4 + 2HF <=> XeO3F2 + H2O}\] This is exactly our target reaction. The new equilibrium constant \(K_3\) is the product of the individual steps: \[K_3 = K'_1 \times K_2\] \[K_3 = \frac{1}{K_1} \times K_2 = \frac{K_2}{K_1}\] Step 5 - Evaluate the Options * **Option (A) \(K_1 / K_2\):** Incorrect. This represents the equilibrium constant for the reverse target reaction. * **Option (B) \(K_1 + K_2\):** Incorrect. Equilibrium constants are multiplied, not added, when reactions are added together. * **Option (C) \(K_2 / K_1\):** Correct. As mathematically demonstrated, \(K_3 = \frac{K_2}{K_1}\). * **Option (D) \(K_2 - K_1\):** Incorrect. Equilibrium constants are never subtracted. \[\boxed{\text{C}}\]