Question

4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of HI were found. The equilibrium constant for H₂(g) + I₂(g) ⇋ 2HI(g) is _____________ .

Answer 1

Correct Answer: 1

Step 1: Write the Balanced Chemical Equation and ICE Table

The balanced chemical equation is: 

\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g). \]

Using the ICE (Initial, Change, Equilibrium) table:

Species	Initial (t = 0)	Change	Equilibrium (t = eq)
\( \text{H}_2 \)	4.5	\(-x\)	\(4.5 - x\)
\( \text{I}_2 \)	4.5	\(-x\)	\(4.5 - x\)
\( \text{HI} \)	0	\(+2x\)	\(2x\)

We are given that the equilibrium moles of \( \text{HI} \) are 3. Thus:

\[ 2x = 3 \implies x = \frac{3}{2} = 1.5. \]

At equilibrium:

• Moles of \( \text{H}_2 \) = \( 4.5 - 1.5 = 3 \)
• Moles of \( \text{I}_2 \) = \( 4.5 - 1.5 = 3 \)
• Moles of \( \text{HI} \) = 3

Step 2: Calculate the Equilibrium Concentrations

Divide the equilibrium moles by the volume of the vessel (10 L):

• \([ \text{H}_2 ] = \frac{3}{10} = 0.3 \, \text{M}\)
• \([ \text{I}_2 ] = \frac{3}{10} = 0.3 \, \text{M}\)
• \([ \text{HI} ] = \frac{3}{10} = 0.3 \, \text{M}\)

Step 3: Calculate the Equilibrium Constant

The equilibrium constant \( K_c \) is given by:

\[ K_c = \frac{[ \text{HI} ]^2}{[ \text{H}_2 ][ \text{I}_2 ]}. \]

Substitute the equilibrium concentrations:

\[ K_c = \frac{(0.3)^2}{(0.3)(0.3)} = \frac{0.09}{0.09} = 1. \]

Final Answer:

The equilibrium constant (\( K_c \)) is 1.