Calculate Kp at 500K and the mole fraction of HI present at 500K and 1atm. What

Question

Calculate Kp at 500K and the mole fraction of HI present at 500K and 1atm. What happens at 10 atm? The answer for Kp is 112.9, mole fraction of HI is 0.842. can anyone provide the steps for the solution? thanks!

2. (Castellan, Problem 11.8) At 500K we have the data One mole of H 2 and one mole of are placed in a vessel at 500K. At this temperature only gases are present and the equilibrium is established. Calculate K at 500K and the mole fraction of HI present at 500K and l atm. What happens at 10 atm?

Explanation / Answer

Kp =[HI]2/ [H][I]

Kp=[10.0-2x]2 /x2 =y^2=131.9

x=0.741 atm

PH2 = 10.0 atm - (2 x 0.7416) atm = 8.517

Since there are equal moles on each side of the equation, the equilibrium constant
does not change when the total pressure is increases. The mole fraction of each gas
is also constant

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