Question

At 171 oC(degrees celsius) the equilibrium constant for the reaction: 2 HBr(g) H2(g) + Br2(g) is KP = 3.45e-11. If the initial pressure of HBr is 0.00819 atm, what are the equilibrium partial pressures of HBr, H2, and Br2?

Answer 1

Kp = PH2*PBr2/ (PHBr)^2

Kp = 3.45*10^-11

note that reactants are favoured, i.e. HBr will react almost completely

if; initially:

PHBr = 0.00819

PH2 = 0

PBr2 = 0

then

in equilibrium, due to stoichiometry and

PHBr = 0.00819 - 2x

PH2 = 0 + x

PBr2 = 0 + x

substitute in Kp expression

Kp = PH2*PBr2/ (PHBr)^2

3.45*10^-11 = x*x/(0.00819 - 2x)^2

solve for x

sqrt(3.45*10^-11) = x/((0.00819 - 2x)

(5.87*10^-6) = x/(0.00819 - 2x)

(0.00819 - 2x) = x/((5.87*10^-6) )

0.00819 -2x = 170251x

170253x = 0.00819

x = 0.00819/170253 = 4.81048*10^-8

meaning that

PHBr = 0.00819 - 2x = 0.00819 - (2* 4.81048*10^-8) = 0.00818990379

PH2 = 0 + x = 4.81048*10^-8

PBr2 = 0 + x = 4.81048*10^-8

whats happening?

no reactions, since the Kp is strongly shifteed toward reactants (HBr) anr will not REACT since products are not favoured