At 603K,Kc for the reaction NH4Cl(g)<==>NH3(g) + HCl(g) is 0.011.What is the equilibrium concentration of NH3(g) if the initial concentration of NH4Cl(g) is 0.500M?

Answer 1

.................NH4Cl ==> NH3 + HCl
I................0.5 M............0..........0
C...............-x..................x...........x
E...............0.5-x.............x............x

Kc = 0.011 = (NH3)(HCl)/NH4Cl
Plug the E line into the Kc expression and solve for x = (NH3).

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