Question

2. In the reaction H2(g) + b(0) + 2HI(), there are equilibrium partial pressures of PH, = 13.0 MPa, P = 11.7 MPa and Phi = 66.6 MPa. What is the
equilibrium constant K, for this reaction?
O A 0.44
B.228
OC.143
D. 292

Answer 1

Answer:

Equilibrium constant is equal to $29.162$

Explanation:

If the reaction is of this form

aA(g) + bB(g) ⇄ cC(g) + dD(g)

Then , the Kp of this equation will be equal to

$K_p =\frac{P_C^cP_D^d}{P_B^bP_A^a}$

Given

$P_c =$$66.6$ MPa and c is equal to two

$P_A = 13.0$ MPa

$P_b = 11.7$ MPa

Here a and b is equal to zero.

Substituting the given values we get -

$K_p = \frac{66.6^2}{13*11.7}\\K_p = 29.162$

Equilibrium constant is equal to $29.162$