Question

For the reaction at \( T(K) \), \( A_2(g) \rightleftharpoons B_2(g) \), \( K_c \) for the reaction at \( T(K) \) is 39.0. In a closed 1L flask, one mole of \( A_2(g) \) was heated to \( T(K) \). What is the concentration of \( B_2(g) \) at equilibrium (in mol L\(^{-1}\))?

• A. 0.975
• B. 0.015
• C. 0.65
• D. 0.035

Hint:

When dealing with equilibrium problems, remember that \( K_c \) expresses the ratio of concentrations of products to reactants raised to the power of their coefficients.

Answer 1

The Correct Option is A

Step 1: Define Initial and Equilibrium Concentrations
Let the initial concentration of \( A_2(g) \) be 1 mol L\(^{-1}\) (since 1 mole is placed in a 1L flask). At equilibrium, let x be the concentration of \( B_2(g) \) formed. Since the reaction is: \[ A_2(g) \rightleftharpoons B_2(g) \] The equilibrium concentrations will be: - \( [A_2] = (1 - x) \) - \( [B_2] = x \)
Step 2: Apply the Equilibrium Constant Expression
The equilibrium constant for the reaction is given by: \[ K_c = \frac{[B_2]}{[A_2]} \] Substituting the values: \[ 39 = \frac{x}{1 - x} \]
Step 3: Solve for \( x \)
Rearranging the equation: \[ x = 39(1 - x) \] \[ x + 39x = 39 \] \[ 40x = 39 \] \[ x = \frac{39}{40} = 0.975 \]
Step 4: Conclusion
Thus, the equilibrium concentration of \( B_2(g) \) is 0.975 mol L\(^{-1}\), so the correct answer is option (1).