Question

A sample of solid NH4Cl was placed in a sealed jar and was heated to 465 °C until the system reached equilibrium. The pressure of HCl and NH3 were 1.03 and 1.18 atm, respectively. Determine the value of ΔG° for the reaction at 465 °C.

Answer 1

Answer:

- 1.22KJ

Explanation:

First we put down the equation of the reaction;

NH4Cl(s)<-------->NH3(g) + HCl(g)

The equilibrium constant K is obtained from;

K= P(NH3) × P(HCl)

Where;

P(NH3)= pressure of ammonia

P(HCl)= pressure of hydrogen chloride gas

Pure solids and liquids are not included in the equilibrium constant expression. This is because they do not affect the reactant amount at equilibrium in the reaction, so they are disregarded and kept at 1. Remember that the activity, a, of any solid or liquid in a reaction is equal to 1. Hence [NH4Cl]= 1

K= 1.03 × 1.18 = 1.22

∆G° = -RTlnK

R= 8.314 JK-1mol-1

T= 465°C +273= 738K

K= 1.22

lnK= ln(1.22)= 0.19885

∆G= -(8.314 ×738 ×0.19885)

∆G= -1220.1 J

∆G= -1.22×10^3 J

∆G= - 1.22KJ