Given that Kc = 82, calculate the equilibrium pressure of H2.

Question

In the reaction below, 4.85 atm each of H₂ and Cl₂ were placed into a 1.00 L flask and allowed to react:

H₂(g) + Cl₂(g) <=> 2 HCl(g)

Given that Kc = 82, calculate the equilibrium pressure of H₂.  Give answr to 2 decimal places.

The correct answer is 0.88

Answer 1

H₂(g) + Cl₂(g) <=> 2HCl(g)

Kp = Kc(RT)^∆n

∆n = change in moles of gas = 2 - 2 = 0 therefore (RT)^o = 1 so temperature isn't needed

Kp = (82)(RT)⁰ = 82

Kp = P(HCl)² / P(H₂)(PCl₂)

H₂(g) + Cl₂(g) <==> 2HCl(g)

4.85.......4.85...............0.........Initial

-x...........-x...................+2x.....Change

4.85-x....4.85-x...........2x........Equilibrium

82 = (2x)² /(4.85-x)²

Taking square roots....

9.055 = 2x/4.85-x

11.055x = 43.92

x = 3.97 and plugging this into the ICE table for H₂ we obtain...

4.85 - 3.97 = 0.88 <--answer