Question

calculate the equilibrium constant k for the reaction: h₂(g) + i₂(g) ⇌ 2 hi(g) at 448°c if the equilibrium concentrations are: h₂ = 0.0700 m, i₂ = 1.07 m, and hi = 1.90 m.
○48.2
○25.4
○0.0207
○0.0394
○none of the choices are correct.

Explanation:

Step1: Write the formula for equilibrium constant

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

Step2: Substitute the given values

Given \([H_{2}]=0.0700M\), \([I_{2}]=1.07M\), \([HI]=1.90M\)
Substitute into the formula: \(K = \frac{(1.90)^{2}}{0.0700\times1.07}\)
First, calculate \((1.90)^{2}=3.61\)
Then, calculate \(0.0700\times1.07 = 0.0749\)
Now, \(K=\frac{3.61}{0.0749}\approx48.2\)

Answer:

A. 48.2