The following question is in the image For the following rea

The following question is in the image

For the following reaction, H2(g) doubleheadarrow 2HF(g), compute: Delta G degree at 25 degree C. K at 25 degree C. K at 225 degree C. Find the degree of dissociation of the reaction at a) 25 degree C; and b) 225 degree C assuming a total pressure of 1 bar. Does this result agree with le Chatelier\'s prediction? Find the degree of dissocation of the reaction at a) 1 bar; and b) 10 bar assuming a temperature of 25 degree C. Does this result agree with le Catelier\'s prediction? You begin this chemical reaction with partial pressures given as: PH2 = 2.0 bar, PF2 = 3.0 bar, and PHF = 5.0 bar. Find: a) Delta G at 25 degree C; and b) the equilibrium configuration of this system at 25 degree C. Beginning with the equilibrium configuration given in #6 above, you add an additional 2.5 bar of HF to the flask. Calculate the new equilibrium configuration. Does your result agree with le Chatelier?

Solution

H2(g) + F2(g) ? 2HF(g)
The equilibrium constant K can be calculated as:
K = [HF]^2/{ [H2]*[F2] } = 0.4^2/(0.05*0.01) = 320

After the addition of 0.330 mole of F2, the F2 concentrationbecomes (before reaction):
[F2] = 0.330mol /5.00L + 0.0100 M = 0.0760M
Assume an additional X(M) of [F2] will be converted to HF at thenew equilibrium. Thus the new concentration of HF is: [HF] =(0.400+2X) M.
...........H2(g) + F2(g) ? 2HF(g), K = 320
Initial: 0.0500..0.0760...0.400
Final: 0.05-X...0.07-X...0.4+2*X
K = 320 = (0.4+2*X)^2 /((0.05-X)*(0.07-X))
or: (0.4+2*X)^2 = 320*(0.05-X)*(0.07-X)
0.16+1.6*X+4*X^2 = 320*(0.0035-0.12*X+X^2)
316*X^2-40*X+0.96 = 0
The solutions to this quadratic equation are:
X1=0.0944 (to be omitted. Do you see why?)
X2 = 0.0322 (M)
So finally:
[H2] = 0.0178M
[F2] = 0.0378M
[HF] = 0.4644M