This reaction has an equilibrium constant of Kp = 2.2 * 106 at 29...

Question

This reaction has an equilibrium constant of Kp = 2.2 * 106 at 298 K.
2 COF2(g) ⇌ CO2(g) + CF4(g)
Calculate Kp for each reaction and predict whether reactants or
products will be favored at equilibrium. c. 2 CO2(g) + 2 CF4(g) ⇌ 4 COF2(g)

Accepted Answer

Hi everyone. So here we're giving the K. P. For the following reaction below. And this is 1.10 times 10 to the 10 at 78.5°C. We were asked what would be the K. P. value if the following equation became the one below. You see that in the first reaction, hydrazine and hatching gas, other reactant while ammonia is a product. But in the second reaction, ammonia is the reactant, while hydrazine and hydrogen gas are the products. This means the second reaction is the reverse of the first reaction. If we reverse the first reaction that K. P becomes K. P equals one Over the K. P Value for the First Reaction, which is 1.10 times 10 to the 10. And this gives us 9.09. I'm sent to the mega 11 as the K. P. For the second reaction. Thanks for watching my video and I hope it was helpful.

This reaction has an equilibrium constant of Kp = 2.2 * 106 at 298 K. 2 COF2(g) ⇌ CO2(g) + CF4(g) Calculate Kp for each reaction and predict whether reactants or products will be favored at equilibrium. c. 2 CO2(g) + 2 CF4(g) ⇌ 4 COF2(g)

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