Given \(K=3.50\) at \(45^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g) $$ and \(K=7.10\) at \(45^{\circ} \mathrm{C}\) for the reaction $$ 2 \mathrm{~A}(g)+\mathrm{D}(g) \rightleftharpoons \mathrm{C}(g) $$ what is the value of \(K\) at the same temperature for the reaction $$ \mathrm{C}(g)+\mathrm{D}(g) \rightleftharpoons 2 \mathrm{~B}(g) $$ What is the value of \(K_{\mathrm{p}}\) at \(45^{\circ} \mathrm{C}\) for the reaction? Starting with \(1.50\) atm partial pressures of both \(\mathrm{C}\) and \(\mathrm{D}\), what is the mole fraction of \(\mathrm{B}\) once equilibrium is reached?

We are given the following two reactions and their equilibrium constants (\(K1 = 3.50\) and \(K2 = 7.10\)): (1) A(g) + B(g) ⇌ C(g) ; \(K_1 = 3.50\) (2) 2A(g) + D(g) ⇌ C(g) ; \(K_2 = 7.10\) We are asked to find the equilibrium constant for the following reaction: (3) C(g) + D(g) ⇌ 2B(g) ; We want to find its \(K_3.\)

We will manipulate the first two reactions to obtain the desired reaction (3): (a) Multiply the reaction (1) by two: 2A(g) + 2B(g) ⇌ 2C(g) ; \(K_{1_2x} = (K_1)^2\) (b) Subtract the reaction (2) from the reaction (a): [2A(g) + 2B(g) ⇌ 2C(g)] - [2A(g) + D(g) ⇌ C(g)] = C(g) + D(g) ⇌ 2B(g) Here, we need to remember that when we subtract reactions, we divide their equilibrium constants. Thus: \(K_3 = \frac{K_{1_2x}}{K_2}\)

Now substitute the known values of \(K_1\) and \(K_2\) to find \(K_3\): \(K_3 = \frac{(3.50)^2}{7.10} = \frac{12.25}{7.10} \approx 1.725\) So the equilibrium constant for the desired reaction (3) is approximately 1.725.

We are given that at 45°C, the partial pressures of both C and D are 1.50 atm. Let us assume that the change in the partial pressure of B is x atm at equilibrium. We can then write the expression for the reaction quotient Q: \(Q = \frac{(1.50-x)^2}{(1.50)(1.50-x)}\) Since the system is at equilibrium, \(Q = K_3\). So: \(1.725 = \frac{(1.50-x)^2}{(1.50)(1.50-x)}\) Solve for x: \(2.5875 - 3.45x + x^2 = 2.25 - 2.25x\) Rearranging the equation, we get: \(x^2 - 1.20x + 0.3375 = 0\) Now use the quadratic formula to find x: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) Using b = 1.20, a = 1, and c = 0.3375: \(x = \frac{-1.20 \pm \sqrt{(1.20)^2-4(0.3375)}}{2} \approx 0.5886\) Since the change in the partial pressure of B at equilibrium is positive, its partial pressure at equilibrium will be: \(P_{\text{B}} = 1.5 - 0.5886 \approx 0.9114 \: \text{atm}\) Finally, we calculate the mole fraction of B at equilibrium: \(\text{mole fraction of B} =\frac{P_{\text{B}}}{P_{\text{total}}}\) Assuming total pressure doesn't change, we get: \(\text{mole fraction of B} = \frac{0.9114}{1.50} \approx 0.6076\) Hence, the mole fraction of B once equilibrium is reached is approximately 0.6076.