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)

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Chemistry

Donaldson

20 February, 08:27

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)

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Jamir Stafford · 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