Question

consider the following reaction:
2hi(g)h₂(g) + i₂(g)
if 1.41 moles of hi, 0.219 moles of h₂, and 0.253 moles of i₂ are at equilibrium in a 18.3 l container at 928 k, the value of the equilibrium constant, kp, is

Explanation:

Step1: Calculate the molar concentrations

The molar concentration \(C=\frac{n}{V}\).
For \(HI\): \(C_{HI}=\frac{1.41\space mol}{18.3\space L}\approx0.07705\space M\)
For \(H_2\): \(C_{H_2}=\frac{0.219\space mol}{18.3\space L}\approx0.01197\space M\)
For \(I_2\): \(C_{I_2}=\frac{0.253\space mol}{18.3\space L}\approx0.01382\space M\)

Step2: Calculate \(K_c\)

The equilibrium constant expression for the reaction \(2HI(g)
ightleftharpoons H_2(g)+I_2(g)\) is \(K_c = \frac{[H_2][I_2]}{[HI]^2}\)
Substitute the values: \(K_c=\frac{0.01197\times0.01382}{(0.07705)^2}\)
\(K_c=\frac{0.0001655}{0.005936}\approx0.0279\)

Step3: Convert \(K_c\) to \(K_p\)

The relationship between \(K_p\) and \(K_c\) is \(K_p = K_c(RT)^{\Delta n}\)
For the reaction \(\Delta n=(1 + 1)-2=0\)
Since \(\Delta n = 0\), \(K_p=K_c\) (because \((RT)^0 = 1\))

Answer:

\(0.0279\)