Question

A(g) + 2B(g) → C(g) + D(g)If you initially start with 1.00 atm of both A and B and find that at equilibrium 0.211 atm of C is present, what is the value of Kp for the reaction at the temperature the reaction was run?

Answer 1

Answer: The value of
`K_p` for the reaction is 0.169

Step-by-step explanation:

We are given:

Initial partial pressure of A = 1.00 atm

Initial partial pressure of B = 1.00 atm

The given chemical equation follows:

`A(g)+2B(g)\rightleftharpoons C(g)+D(g)`

Initial: 1.00 1.00

At eqllm: 1-x 1-2x x x

We are given:

Equilibrium partial pressure of C = 0.211 atm = x

So, equilibrium partial pressure of A = (1.00 - x) = (1.00 - 0.211) = 0.789 atm

Equilibrium partial pressure of B = (1.00 - 2x) = (1.00 - 2(0.211)) = 0.578 atm

Equilibrium partial pressure of D = x = 0.211 atm

The expression of
`K_p` for above equation follows:

`K_p=(p_C* p_D)/(p_A* (p_B)^2)`

Putting values in above equation, we get:

`K_p=(0.211* 0.211)/(0.789* (0.578)^2)\\\\K_p=0.169`

Hence, the value of
`K_p` for the reaction is 0.169