Question

At 730°C the Kc for the reaction H2(g)+ Br2(g)<--->2HBr(g) is 2.18x10^6. If 3.25mol HBr(g) is placed in a 12.0 L reaction vessel at this temperature, how many moles of each of the three gases will be present at equilibrium?

Answer 1

H2(g)+ Br2(g) <---------> 2HBr(g)

0              0                      3.25

x               x                     3.25 - x

Kc = [HBr]^2 / [H2][Br2]

2.18x10^6 = (3.25 - x)^2 / x^2

3.25 - x / x = 1476.5

x = 2.20 x 10^-3

moles of each gas at equilibrium :

moles of H2 = 0.00220

moles of Br2 = 0.00220

moles of HBr = 3.248 mol