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=((P_(HI))^2)/(P_(H_2)* 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=((52.3)^2)/(61.8* 46.5)\\\\K_p=0.952`

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