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

To find the equilibrium concentrations of \(A(g)\) and \(B(g)\), we need to analyze the reaction \[ A(g) \rightleftharpoons T(K) + B(g) \] with a given equilibrium constant (\(K_c\)) of 39.0.

Initially, 1 mole of \(A(g)\) is present in a 1-litre flask at \(T(K)\), making the initial concentration of \(A(g)\), \([A]_0 = 1 \, \text{mol L}^{-1}\).

At equilibrium, let the change in concentration of \(A(g)\) that reacted be \(x \, \text{mol L}^{-1}\). Therefore, the equilibrium concentrations can be expressed as:

• \([A] = 1 - x\)
• \([B] = x\)

Substituting these expressions into the equilibrium expression for \(K_c\):

$$\frac{\mathrm{[B]}}{\mathrm{[A]}}=\; K\_c$$

Thus,

$\frac{x}{\mathrm{1-x}}=\; 39.0$

Solving for \(x\):

\(x = 39.0(1 - x)\)

\(x = 39.0 - 39.0x\)

\(x + 39.0x = 39.0\)

\(40.0x = 39.0\)

\(x = \frac{39.0}{40.0}\)

\(x = 0.975\)

This implies:

• At equilibrium, \([B] = x = 0.975 \, \text{mol L}^{-1}\)
• \([A] = 1 - x = 0.025 \, \text{mol L}^{-1}\)

Thus, the equilibrium concentrations of \(A(g)\) and \(B(g)\) are respectively 0.025 mol L\(^{-1}\) and 0.975 mol L\(^{-1}\). The correct answer is \(0.025, 0.975\).

Answer 2

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 \( [B(g)] = 0.025 \, {mol/L} \).