Question:
The equilibrium constant for the dissociation of HI at 773 K is:
The equilibrium constant for the dissociation of HI at 773 K is:
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Equilibrium constant is calculated using molar concentrations at equilibrium.
Updated On: Mar 24, 2025
- \( 2 \times 10^{-2} \)
- \( 50 \)
- \( 2 \times 10^{-1} \)
- \( 5.0 \)
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The Correct Option is A
Solution and Explanation
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:
\[
K_c = 2 \times 10^{-2}
\]
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