Question

The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) you may want to reference (pages 680 - 684) section 15.3 while completing this problem. part a calculate kp for the reaction below. 4cof2(g)⇌2co2(g)+2cf4(g)

Answer 1

Kp is the equilibrium pressure constant calculated from the partial pressures of a reaction equation.
 Kp =[pCF4]*[p CO2] / [p COF2]^2 = 2.2 x 10^6
When the mole fraction gets doubled we have
 Kp = [pCO2]^2*[pCF4]^2 / [pCOF2]^4
 Kp = [[pCF4]*[p CO2] / [p COF2]^2] * 2
Kp = (2.2 * 10^6) * 2
Kp = 4.8 * 10^12