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

D. 29.2

Step-by-step explanation:

To calculate the equilibrium constant Kp for an equation of the form aA + bB ⇌ cC, use the following formula: Kp = (PC)c(PA)a(PB)b

For the equation H2(g) + I2(g) ⇌ 2HI(g), we have:

Kp= PHI2PH2 ⋅ PI2 = 66.6213.0 ⋅ 11.7 = 29.2

Marked correct on Gizmo :)

Answer 2

Equilibrium constant is equal to
`29.162`

Step-by-step 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 =(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 = (66.6^2)/(13*11.7)\\K_p = 29.162`

Equilibrium constant is equal to
`29.162`