Answer: The value of [tex]K_p[/tex] for the reaction is 0.169
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:
[tex]A(g)+2B(g)\rightleftharpoons C(g)+D(g)[/tex]
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 [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_C\times p_D}{p_A\times (p_B)^2}[/tex]
Putting values in above equation, we get:
[tex]K_p=\frac{0.211\times 0.211}{0.789\times (0.578)^2}\\\\K_p=0.169[/tex]
Hence, the value of [tex]K_p[/tex] for the reaction is 0.169
