Question

The equilibrium constant for the dissociation of HI at 773 K is:

• A. \( 2 \times 10^{-2} \)
• B. \( 50 \)
• C. \( 2 \times 10^{-1} \)
• D. \( 5.0 \)

Hint:

Equilibrium constant is calculated using molar concentrations at equilibrium.

Answer 1

The Correct Option is A

Step 1: Write the Reaction \[ 2HI \rightleftharpoons H_2 + I_2 \] Step 2: Apply Equilibrium Expression \[ K_c = \frac{[H_2][I_2]}{[HI]^2} \] Solving: \[ K_c = 2 \times 10^{-2} \]