Question

Hydrogen and iodine react to form hydrogen iodide, like this: H_2 (g) + I_2 (g) rightarrow 2 HI(g) Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition: Calculate the value of the equilibrium constant K_p for this reaction. Round your answer to 2 significant digits.

Answer 1

This is an incomplete question, here is a complete question.

Hydrogen and iodine react to form hydrogen iodide, like this:

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen, iodine, and hydrogen iodide has the following composition:

Compound            Pressure at equilibrium

$H_2$                                   61.8 atm

$I_2$                                    46.5 atm

$HI$                                  52.3 atm

Calculate the value of the equilibrium constant $K_p$ for this reaction. Round your answer to 2 significant digits.

Answer : The value of equilibrium constant $K_p$ for this reaction is, 0.952

Explanation :

The given chemical reaction :

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{(P_{HI})^2}{P_{H_2}\times P_{I_2}}$

We are given:

$P_{H_2}=61.8atm$

$P_{I_2}=46.5atm$

$P_{HI}=52.3atm$

Putting values in above equation, we get:

$K_p=\frac{(52.3)^2}{61.8\times 46.5}\\\\K_p=0.952$

Therefore, the value of equilibrium constant $K_p$ for this reaction is, 0.952