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: \[ K_c = 2 \times 10^{-2} \]

Answer 2

The dissociation of hydrogen iodide (HI) can be represented by the equilibrium reaction:

\[ 2 HI (g) \rightleftharpoons H_2 (g) + I_2 (g) \]

The equilibrium constant expression \( K_c \) for this reaction is given by:
\[ K_c = \frac{[H_2][I_2]}{[HI]^2} \]

At 773 K, the value of the equilibrium constant \( K_c \) for the dissociation of HI is experimentally determined and is known to be:
\[ K_c = 2 \times 10^{-2} \]

This relatively small value indicates that, at 773 K, the equilibrium lies significantly toward the reactants (HI), meaning only a small fraction of HI dissociates into H₂ and I₂ at this temperature.

Therefore, the correct value of the equilibrium constant for the dissociation of HI at 773 K is:
\[ \boxed{2 \times 10^{-2}} \]