Question

± Pressure-Based versus Concentration-Based Equilibrium Constants

Part A.) For the reaction

3A(g)+3B(g)⇌C(g)

Kc = 67.6 at a temperature of 287 ∘C .

Calculate the value of Kp.

Part B.) For the reaction

X(g)+3Y(g)⇌3Z(g)

Kp = 1.65×10⁻² at a temperature of 209 ∘C .

Calculate the value of Kc.

Answer 1

Part A:

3A(g)+3B(g)⇌C(g)

Given K_c = 67.6

T = temperature = 287 ∘C = 287+273 = 560 K

We know that K_p = K_cx (RT) ⁿ

Where

R = gas constant = 0.0821Latm/(mol-K)

n = change in number of moles

      = total number of moles of gaseous products - total number of moles of gaseous reactants

      = 1 - (3+3)

      = -5

Plug the values we get

K_p = K_cx (RT) ⁿ

K_p = 67.6x (0.0821x560) ^-5

    = 67.6x40.976

    = 2.770x10³

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Part B:

X(g)+3Y(g)⇌3Z(g)

Given K_p = 1.65x10^-2

T = temperature = 209 ∘C = 209+273 = 482 K

We know that K_p = K_cx (RT) ⁿ

Where

R = gas constant = 0.0821Latm/(mol-K)

n = change in number of moles

      = total number of moles of gaseous products - total number of moles of gaseous reactants

      = 3 - (1+3)

      = -1

Plug the values we get

K_c = K_px (RT) ^-n

K_c = 1.65x10^-2x (0.0821x482) ¹

    = 1.65x10^-2x39.57

    = 0.653