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 information: 1. For the reaction: \(A(g) + B(g) \rightleftharpoons C(g)\), we have \(K_1 = 3.50\) at \(45^\circ \mathrm{C}\) 2. For the reaction: \(2A(g) + D(g) \rightleftharpoons C(g)\), we have \(K_2 = 7.10\) at \(45^\circ \mathrm{C}\) 3. We need to find \(K_3\), the equilibrium constant for the reaction: \(C(g) + D(g) \rightleftharpoons 2B(g)\)

First, we will multiply the first reaction by 2, to make the number of A's equal to the second reaction, resulting in: \[2A(g) + 2B(g) \rightleftharpoons 2C(g) \quad (1')\] Now, we can subtract the second reaction from the first reaction (1'): \((1') - (2) \Rightarrow )[2A(g) + 2B(g) \rightleftharpoons 2C(g)] - [2A(g) + D(g) \rightleftharpoons C(g)]\] Which results in the target reaction: \[C(g) + D(g) \rightleftharpoons 2B(g)\]

When we multiplied the first reaction by 2, the new equilibrium constant becomes \(K_1^{'} = K_1^2\), so we get: \(K_1^{'} = 3.50^2 = 12.25\) When we subtracted the second reaction from the first reaction, the new equilibrium constant will be the quotient of the two constants \(K_3 = \frac{K_1^{'}}{K_2}\), so we have: \(K_3 = \frac{12.25}{7.10} = 1.725\) So, the value of \(K_3\) at the same temperature for the reaction \(C(g) + D(g) \rightleftharpoons 2B(g)\) is 1.725.

We are given that the initial partial pressures of C and D are both 1.50 atm. Let the change in pressure of B at equilibrium be denoted as \(x\). Then, the pressure of B at equilibrium will be \(2x\), and the pressure of C and D will be \(1.50 - x\) atm each. The reaction quotient, \(Q_p\), for the target reaction can be written as: \(Q_p = \frac{P^2_{B}}{P_{C} \cdot P_{D}}\) At equilibrium, \(Q_p\) is equal to \(K_p\). Thus, \(K_p = \frac{(2x)^2}{(1.50 - x)(1.50 - x)}\) We are given that \(K_p = K_3 = 1.725\), so: \(1.725 = \frac{4x^2}{(1.50 - x)^2}\) Now, we will solve for \(x\). Since we are looking for the mole fraction of B, which must be less than 1, we can discard any extraneous solutions. \(x \approx 0.468\) The mole fraction of B at equilibrium is: \[\frac{2x}{1.50 + 1.50} = \frac{2(0.468)}{3.00} \approx 0.312\] Thus, the mole fraction of B once equilibrium is reached is approximately 0.312.