Question

CaCO3 (s) ⇌ CaO (s) + CO2 (g)

At 637 °C, the reaction reaches equilibrium.

If P_CO2 =0.89 atm, what is Kc value ?

Answer 1

Equilibrium constant (Kp) is the product of the pressure of gaseous products divided by the product of the pressure of gaseous reactant with each term raised to their coefficients as power.

Kp = PCO2 = 0.89

Now, Kp = Kc(RT)∆n

Where ∆n = moles of gaseous product - moles of gaseous reactant

= 1-0 = 1

T = 637°C = 637+273.15 = 910.15 K

R = 0.083 Latm/mol.K

Hence, 0.89 = Kc(0.082×910.15)

Kc = 0.89/0.082×910.15

= 0.012