Calculate H, S, and G at 298 K for each ofthe following reactions. In each case

Question

Calculate H, S, and G at 298 K for each ofthe following reactions. In each case show thatG=H-TS. A) H2(g) + F2(g)--->2HF(g) B) C(s graphite) + 2Cl2(g) ---->CCl4(g) C) 2PCl3(g) + O2(g)---->2POCl3(g) D) 2CH3OH(g) + H2(g)-----> C2H6(g) +2H2O(g) Calculate H, S, and G at 298 K for each ofthe following reactions. In each case show thatG=H-TS. A) H2(g) + F2(g)--->2HF(g) B) C(s graphite) + 2Cl2(g) ---->CCl4(g) C) 2PCl3(g) + O2(g)---->2POCl3(g) D) 2CH3OH(g) + H2(g)-----> C2H6(g) +2H2O(g) D) 2CH3OH(g) + H2(g)-----> C2H6(g) +2H2O(g)

Explanation / Answer

a) 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

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