Calculate Kc for the reaction below?

I2 (g) <—–> 2I (g) kp=6.26*10^-22 At 298 K

Had the exact problem in Mastering Chem!!

So the formula that you use is Kp = Kc(RT)^Δn

Since you’re working with a gas, use the ideal gas law constant: R = 0.08205

T is already in Kelvin

For Δn, subtract the number of moles of your product from your reactant side: 2-1=1

Then plug it all in: (6.26*10^-22) = Kc (0.08205 * 298)^1

6.26*10^-22 = Kc*24.4509

Kc = 2.56*10^ -23
For situation #2, use the equation Kp=Kc(RT)^delta n by using fact you have 2 moles on the two aspects of the equation, delta n would be 0, subsequently making your equation for Kp=(80 one.9)/(.08206*298)^0) answer is 80 one.9