Question

For the reaction 2 SO2(g) + O2(g) →→2 SO3(g) AG° = -140.3 kJ and AS = -187.9 J/K at 306 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 306 K. The standard enthalpy change for the reaction of 1.79 moles of so,(g) at this temperature would be

Answer 1

Solution:

Part A)

For the given reaction,

2SO2 (g) + O2 (g) = 2SO3 (g)

ΔG° = - 140.3 kJ at 306 K and 1 atm

Since, ΔG° = - ve, hence process is spontaneous and therefore reaction favoures formation of product.

Part B)

The standard free energy change (ΔG°) related with ΔH° and ΔS° as,

ΔG° = ΔH° - T ΔS°

ΔH° = ΔG° + TΔS°

= -140.3 kJ + 306 K x -187.9 J K-1

= - 140.3 KJ - 57405.6 J

= - 140.3 kJ - 57.4056 KJ

= -197.71 KJ

For 2 mol of SO2 enthalpy change = -197.71 KJ

For 1.79 mol = 1.79 x -197.71 KJ /2 = -176.95 KJ