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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

Respuesta :

Answer:

Equilibrium constant is equal to [tex]29.162[/tex]

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

[tex]K_p =\frac{P_C^cP_D^d}{P_B^bP_A^a}[/tex]

Given

[tex]P_c =[/tex][tex]66.6[/tex] MPa and c is equal to two

[tex]P_A = 13.0[/tex] MPa

[tex]P_b = 11.7[/tex] MPa

Here a and b is equal to zero.

Substituting the given values we get -

[tex]K_p = \frac{66.6^2}{13*11.7}\\K_p = 29.162[/tex]

Equilibrium constant is equal to [tex]29.162[/tex]

starzellas8

Answer:

D. 29.2

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 :)

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