# How to calculate the Kp of this equilibrium constant?

The reaction below has an equilibrium constant Kp= 2.2*10^6 at 298 K.

2COF2 (g) –>CO2 (g) + CF4 (g)

Calculate Kp for the reaction below.

2/3COF2 (g) –> 1/3CO2 (g) + 1/3CF4 (g)

Calculate Kp for the reaction below.

4COF2 (g) –> 2CO2 (g) + 2CF4 (g)

Calculate Kp for the reaction below.

CO2 (g) + CF4 (g) –> 2COF2 (g)

All answers have to be in two sig figs please explain thanks

1. The new reaction is 1/3 of the original reaction. So Kp new = (Kp old)^1/3 or 1 / (Kp old)^3 (the cube root of Kp old) = (2.2 x 10^6)^1/3 = 130

2. The new reaction is 2 times the original. Kp new = (Kp old)^2 = (2.2 x 10^6)^2 = 4.8 x 10^12

3. The new reaction is the reverse of the original. Kp new = 1 / Kp old = 1 / (2.2 x 10^6) = 4.5 x 10^-7

You might say that you multiplied the original by -1 (to reverse it). Kp new = (Kp old)^-1 which is the same as the inverse 1 / Kp old.

HPV..You made an ERROR in the third solution.

Here is the right one

Kp new = 1 / Kp old = 1 / (4.8 x 10^12) = 2.1 x 10^-13

Chemistry Student ðŸ™‚

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