Question

consider the following reaction:
2hi(g) ⇌ h₂(g) + i₂(g)
if 1.72 moles of hi(g), 0.248 moles of h₂, and 0.334 moles of i₂ are at equilibrium in a 18.3 l container at 928 k, the value of the
equilibrium constant, kc, is
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Explanation:

Step1: Calculate molar concentrations

The formula for molar concentration \(c=\frac{n}{V}\).
For \(HI\): \(c_{HI}=\frac{1.72\space mol}{18.3\space L}\approx0.094\space M\)
For \(H_{2}\): \(c_{H_{2}}=\frac{0.248\space mol}{18.3\space L}\approx0.0136\space M\)
For \(I_{2}\): \(c_{I_{2}}=\frac{0.334\space mol}{18.3\space L}\approx0.0183\space M\)

Step2: Write the equilibrium constant expression

For the reaction \(2HI(g)
ightleftharpoons H_{2}(g)+I_{2}(g)\), the equilibrium constant \(K_{c}=\frac{[H_{2}][I_{2}]}{[HI]^{2}}\)

Step3: Substitute the values into the expression

\(K_{c}=\frac{0.0136\times0.0183}{(0.094)^{2}}\)
First, calculate the numerator: \(0.0136\times0.0183 = 0.00024888\)
Then, calculate the denominator: \((0.094)^{2}=0.008836\)
\(K_{c}=\frac{0.00024888}{0.008836}\approx0.0282\)

Answer:

\(0.0282\)