Question

15. At 699 K AG° = -23.25 kJ for the reaction H2g) + 12(a)= 2Hg). Calculate AS this reaction if the reagents are both supplied at 10.0 atm pressure and the pro is at 1.94 atm pressure A) -19.1 kJ B) 19.1 kJ C) 42 kJ D) 42 3 kJ E) 42 3 kJ

Answer 1

Solution :-

H2(g) + I2(g) ---- > 2 HI(g)

Delta Go = -23.25 kJ = -23250 J

Temperature = 699 K

Pressure of reactant and products

[H2]=[I2] = 10.0 atm

[HI] = 1.94 atm

Formula to calculate the delta G

Delta G= delta Go – RT ln Q

Where Q= [product]/[reactant]

Lets calculate the value of Q

Q= [HI]^2/[H2][I2]

Q=[1.94]^2/[10.0][10.0]

Q= 0.03764

Lets put the value of Q in the above formula

Delta G= delta Go + RT ln Q

             = -23250 J + (8.314 J per mol K * 699 K * ln (0.03764))

            = -42310 J

Lets convert joules to kJ

-42310 J * 1 kJ / 1000 J = -42.3 kJ

Therefore the answer is option D