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10 moles of A(2), 15 moles of B(2) and 5...

10 moles of `A_(2), 15` moles of `B_(2)` and 5 moles of AB are placed in a 2 litre vessel and allowed the come to equilibrium. The final concentration of AB is 10.5 M, `A_(2)(g)+B_(2)(g)hArr 2AB(g)`
Determine the value of equilibrium constant `(K_(C))` for the reaction.

A

25.3

B

31.5

C

36.3

D

40.5

Text Solution

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The correct Answer is:
To determine the equilibrium constant \( K_C \) for the reaction \[ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \] we start by analyzing the initial conditions and changes that occur during the reaction. ### Step 1: Initial Moles and Concentrations Given: - Initial moles of \( A_2 = 10 \) moles - Initial moles of \( B_2 = 15 \) moles - Initial moles of \( AB = 5 \) moles - Volume of the vessel = 2 L We can calculate the initial concentrations: \[ \text{Concentration of } A_2 = \frac{10 \text{ moles}}{2 \text{ L}} = 5 \text{ M} \] \[ \text{Concentration of } B_2 = \frac{15 \text{ moles}}{2 \text{ L}} = 7.5 \text{ M} \] \[ \text{Concentration of } AB = \frac{5 \text{ moles}}{2 \text{ L}} = 2.5 \text{ M} \] ### Step 2: Change in Concentration at Equilibrium Let \( x \) be the number of moles of \( A_2 \) and \( B_2 \) that react at equilibrium. The changes in moles will be: - Moles of \( A_2 \) at equilibrium = \( 10 - x \) - Moles of \( B_2 \) at equilibrium = \( 15 - x \) - Moles of \( AB \) at equilibrium = \( 5 + 2x \) ### Step 3: Equilibrium Concentration of AB We are given that the final concentration of \( AB \) is \( 10.5 \) M. Therefore, we can set up the equation: \[ \frac{5 + 2x}{2} = 10.5 \] ### Step 4: Solve for \( x \) Multiplying both sides by 2: \[ 5 + 2x = 21 \] Subtracting 5 from both sides: \[ 2x = 16 \] Dividing by 2: \[ x = 8 \] ### Step 5: Calculate Equilibrium Concentrations Now we can find the equilibrium concentrations: - Concentration of \( A_2 \): \[ \frac{10 - x}{2} = \frac{10 - 8}{2} = \frac{2}{2} = 1 \text{ M} \] - Concentration of \( B_2 \): \[ \frac{15 - x}{2} = \frac{15 - 8}{2} = \frac{7}{2} = 3.5 \text{ M} \] - Concentration of \( AB \): \[ \frac{5 + 2x}{2} = \frac{5 + 16}{2} = \frac{21}{2} = 10.5 \text{ M} \] ### Step 6: Calculate the Equilibrium Constant \( K_C \) The equilibrium constant \( K_C \) is given by: \[ K_C = \frac{[AB]^2}{[A_2][B_2]} \] Substituting the equilibrium concentrations: \[ K_C = \frac{(10.5)^2}{(1)(3.5)} = \frac{110.25}{3.5} \] Calculating \( K_C \): \[ K_C = 31.5 \] ### Final Answer The value of the equilibrium constant \( K_C \) is \( 31.5 \). ---
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