Question

49. At a temperature of 298.15 Kelvin, the free
energy change of a reaction (G°) is 19.737
kCal/mole. If the universal gas constant (R) =
1.98717 Calorie/degree/mole, the log10 of the
equilibrium constant K is
(rounded
off to two decimal places).​

Answer 1

Explanation:

Br

2

(l)+Cl

2

(g)→2BrCl(g),ΔH

o

=29.3kJ

S

Br

=152.3S

Cl

2

(g)

=223.0

S

BrCl(g)

=239.7Jmol

1

K

1

The entropy change for the reaction is

ΔS

R

=2S

BrCl(g)

−[S

Br

+S

Cl

2

(g)

]=2×239.7−[223.0+152.3]=104.4Jmol

1

K

1

The relationship between the free energy change, the enthalpy change and the entropy change is as shown.

Δ

r

G=ΔH−TΔS

Substitute values in the above expression.

Δ

r

G=29300298×104.4=1721.8J

Hence, the free energy change for the reaction is 1721.8J