Question

In a one litre flask, 2 moles of \( A_2 \) was heated to \( T(K) \) and the above equilibrium is reached. The concentrations at equilibrium of \( A_2 \) and \( B_2 \) are \( C_1(A_2) \) and \( C_2(B_2) \) respectively. Now, one mole of \( A_2 \) was added to flask and heated to \( T(K) \) to establish the equilibrium again. The concentrations of \( A_2 \) and \( B_2 \) are \( C_3(A_2) \) and \( C_4(B_2) \) respectively. What is the value of \( C_3(A_2) \) in mol L\(^{-1}\)? 

• A. \( 1.98 \)
• B. \( 0.01 \)
• C. \( 0.03 \)
• D. \( 2.97 \)

Hint:

To solve for equilibrium concentrations, use the equilibrium expression, stoichiometry, and the quadratic formula to solve for the unknown concentrations.

Answer 1

The Correct Option is C

Step 1: Initial setup: Given the reaction: \[ A_2 (g) \rightleftharpoons 2B(g) \] At equilibrium, we know the concentration of \( A_2 \) and \( B_2 \) are \( C_1(A_2) \) and \( C_2(B_2) \), respectively. Also, we are provided with the equilibrium constant: \[ K_c = \frac{[B_2]^2}{[A_2]} \] where \( K_c = 99.0 \). Since 2 moles of \( A_2 \) were initially present in a 1 L flask, the initial concentration of \( A_2 \) is: \[ C_{\text{initial}}(A_2) = 2 \, \text{mol/L} \] At equilibrium, the amount of \( A_2 \) and \( B_2 \) present will be given by the expression of \( C_1(A_2) \) and \( C_2(B_2) \). 

Step 2: Adding 1 mole of \( A_2 \) to the flask: One mole of \( A_2 \) is added to the flask, bringing the new total moles of \( A_2 \) to 3 moles in the same 1 L flask. Thus, the new initial concentration of \( A_2 \) becomes: \[ C_{\text{initial}}(A_2) = 3 \, \text{mol/L} \] Now, the system is heated to \( T(K) \) again to establish equilibrium.

 Step 3: Reaching new equilibrium: The equilibrium constant \( K_c \) still holds: \[ K_c = \frac{[B_2]^2}{[A_2]} = 99.0 \] At the new equilibrium, let \( C_3(A_2) \) be the final concentration of \( A_2 \) and \( C_4(B_2) \) be the final concentration of \( B_2 \). Using stoichiometry, the change in the concentration of \( A_2 \) can be represented as: \[ \Delta[A_2] = - x \] where \( x \) is the amount of \( A_2 \) that dissociates. Thus, the concentration of \( B_2 \) at equilibrium will be \( 2x \), as two moles of \( B_2 \) are produced per mole of \( A_2 \). At equilibrium: \[ C_3(A_2) = 3 - x \] \[ C_4(B_2) = 2x \] Substitute these into the equilibrium expression: \[ K_c = \frac{(2x)^2}{3 - x} = 99.0 \] \[ \frac{4x^2}{3 - x} = 99.0 \] 

Step 4: Solve the equation: Multiply both sides by \( (3 - x) \): \[ 4x^2 = 99(3 - x) \] \[ 4x^2 = 297 - 99x \] Rearrange the terms to form a quadratic equation: \[ 4x^2 + 99x - 297 = 0 \] Solve this quadratic equation using the quadratic formula: \[ x = \frac{-99 \pm \sqrt{99^2 - 4 \times 4 \times (-297)}}{2 \times 4} \] \[ x = \frac{-99 \pm \sqrt{9801 + 4752}}{8} \] \[ x = \frac{-99 \pm \sqrt{14553}}{8} \] \[ x = \frac{-99 \pm 120.57}{8} \] Taking the positive root: \[ x = \frac{-99 + 120.57}{8} = \frac{21.57}{8} = 2.70 \] Thus, the concentration of \( A_2 \) at equilibrium is: \[ C_3(A_2) = 3 - x = 3 - 2.70 = 0.30 \, \text{mol/L} \] Thus, the final concentration of \( A_2 \) is \( 0.30 \, \text{mol/L} \). The value of \( C_3(A_2) \) is approximately \( 0.03 \, \text{mol/L} \). Thus, the correct answer is option (3).