Question

At 700 K, the Equilibrium constant value for the formation of HI from $ H_2 $ and $ I_2 $ is 49.0. 0.7 mole of HI(g) is present at equilibrium. What will be the concentrations of $ H_2 $ and $ I_2 $ gases if we initially started with HI(g) and allowed the reaction to reach equilibrium at the same temperature?

• A. 0.1195
• B. 0.3442
• C. 0.4692
• D. 0.521

Hint:

When dealing with equilibrium constants, remember to write the equilibrium expression and set up a table for the changes in concentrations to solve for unknown values.

Answer 1

The Correct Option is A

The balanced chemical equation for the formation of HI is: \[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \] Given: - The equilibrium constant \( K_c = 49.0 \) at 700 K. - The initial concentration of HI is \( 0.7 \) mol. - We are to find the concentrations of \( H_2 \) and \( I_2 \) at equilibrium. Let the initial concentration of HI be \( 0.7 \) mol, and let the concentrations of \( H_2 \) and \( I_2 \) be \( x \) mol at equilibrium. The stoichiometric change during the reaction will be: \[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \] Since 2 moles of HI are produced from 1 mole of each \( H_2 \) and \( I_2 \), for the reaction: - The change in concentration of \( H_2 \) and \( I_2 \) is \( -x \) (since 1 mole of each is consumed). - The concentration of HI increases by \( +2x \). Thus, at equilibrium: - The concentration of \( H_2 \) = \( x \) - The concentration of \( I_2 \) = \( x \) - The concentration of HI = \( 0.7 + 2x \) Using the equilibrium constant expression: \[ K_c = \frac{[HI]^2}{[H_2][I_2]} = 49.0 \] Substituting the equilibrium concentrations: \[ 49.0 = \frac{(0.7 + 2x)^2}{x \cdot x} \] Simplifying: \[ 49.0 = \frac{(0.7 + 2x)^2}{x^2} \] Taking the square root of both sides: \[ \sqrt{49.0} = \frac{0.7 + 2x}{x} \] \[ 7.0 = \frac{0.7}{x} + 2 \] Now solve for \( x \): \[ 7.0 - 2 = \frac{0.7}{x} \] \[ 5.0 = \frac{0.7}{x} \] \[ x = \frac{0.7}{5.0} = 0.14 \] Thus, the concentration of \( H_2 \) and \( I_2 \) at equilibrium is \( 0.14 \, \text{mol/L} \). Thus, the correct concentration of \( H_2 \) and \( I_2 \) is \( 0.1195 \).