Question

The partial pressures of the gases when at equilibrium are found to be: A: 6.70 atm, B: 10.1 atm, C: 3.60 atm. Evaluate Kp for this reaction. 2A(g) B(g) --> C(g)

Answer 1

Answer : The value of equilibrium constant (Kp) is,
`7.94* 10^(-3)`

Explanation : Given,

Partial pressure of A at equilibrium = 6.70 atm

Partial pressure of B at equilibrium = 10.1 atm

Partial pressure of C at equilibrium = 3.60 atm

The given chemical reaction is:

`2A(g)+B(g)\rightarrow C(g)`

The expression for equilibrium constant is:

`K_p=((p_(C)))/((p_(A))^2* (p_(B)))`

Now put all the given values in this expression, we get:

`K_p=((3.60))/((6.70)^2* (10.1))`

`K_p=7.94* 10^(-3)`

Thus, the value of equilibrium constant (Kp) is,
`7.94* 10^(-3)`