Question

K_c for the reaction \[ A(g) \rightleftharpoons T(K) + B(g) \] is 39.0. In a closed one-litre flask, one mole of \( A(g) \) was heated to \( T(K) \). What are the concentrations of \( A(g) \) and \( B(g) \) (in mol L\(^{-1}\)) respectively at equilibrium?

• A. \( 0.025, 0.975 \)
• B. \( 0.975, 0.025 \)
• C. \( 0.05, 0.95 \)
• D. \( 0.02, 0.98 \)

Hint:

To solve equilibrium problems, first express the concentrations in terms of a variable for the change in concentration, then substitute into the equilibrium expression to solve for the unknown.

Answer 1

The Correct Option is A

Given the reaction: \[ A(g) \rightleftharpoons T(K) + B(g), \] the equilibrium constant \( K_c \) is given as 39.0.
Initially, we have one mole of \( A(g) \) in a 1 L flask, so the initial concentration of \( A(g) \) is: \[ [A]_{{initial}} = 1 \, {mol/L}. \] Let \( x \) be the amount of \( A(g) \) that dissociates at equilibrium. Therefore, at equilibrium, the concentration of \( A(g) \) will be \( 1 - x \), and the concentration of both \( T(K) \) and \( B(g) \) will be \( x \). From the equilibrium expression: \[ K_c = \frac{[T(K)][B(g)]}{[A(g)]}, \] we know that: \[ 39.0 = \frac{x \cdot x}{1 - x}. \] Solving for \( x \), we get: \[ 39.0 = \frac{x^2}{1 - x}, \] \[ 39.0(1 - x) = x^2, \] \[ 39.0 - 39.0x = x^2, \] \[ x^2 + 39.0x - 39.0 = 0. \] Solving this quadratic equation, we get: \[ x = 0.025. \] Thus, the concentrations at equilibrium are: \[ [A(g)] = 1 - 0.025 = 0.975 \, {mol/L}, \] \[ [B(g)] = 0.025 \, {mol/L}. \] Thus, the correct answer is \( [A(g)] = 0.975 \, {mol/L} \) and \( [B(g)] = 0.025 \, {mol/L} \).