Sulfuryl dichloride may be formed from the reaction of sulfur dioxide and chlorine. SO2(g) + Cl2(g) → SO2Cl2(g) Substance: SO2(g) Cl2(g) SO2Cl2(g) ΔH°f (kJ/mol) at 298 K –296.8 0 –364.0 ΔG°f (kJ/mol) at 298 K –300.1 0 –320.0 S°(J/K • mol) at 298 K 248.2 223.0 311.9 What is ΔG°rxn for this reaction at 600 K?

ΔG°rxn = 28.4kJ/mol at 600K

Explanation:

Using Hess's law, you can find the ΔH°rxn and S°  subtracting ΔH°f of products - ΔH°f of reactants ×its coefficients. In the same way for S°rxn

For example, for the reaction:

aA + bB → cC:

ΔH°rxn = c×ΔH°fC - (a×ΔH°fA + b×ΔH°fB).

S°rxn = c×S°fC - (a×S°fA + b×S°fB).

For the reaction:

SO₂(g) + Cl₂(g) → SO₂Cl₂(g)

ΔH°rxn = 1×ΔH°f{SO₂Cl₂} - (1×ΔH°fSO₂ + 1×ΔH°fCl₂).

S°rxn = 1×S°f{SO₂Cl₂} - (1×S°fSO₂ + 1×S°fCl₂).

As at 298K:

ΔH°f{SO₂Cl₂} = -364.0kJ/mol

ΔH°f{SO₂} = -296.8kJ/mol

ΔH°f{Cl₂} = 0kJ/mol

ΔH°rxn = 1×{-364.4kJ/mol} - (1×-296.8kJ/mol + 1×0).

ΔH°rxn = -67.2kJ/mol at 298K.

S°f{SO₂Cl₂} = 311.9J/molK

S°f{SO₂} = 248.2J/molK

S°f{Cl₂} = 223.0J/molK

S°rxn = 1×{311.9J/molK} - (1×248.2J/molK + 1×223.0J/molK).

S°rxn = -159.3J/molK = -0.159.3KJ/molK

Using:

ΔG°rxn = ΔH°rxn - S°rxn×T

Assuming ΔH°rxn doesn't change at 600K:

ΔG°rxn = -67.2kJ/mol - -0.159.3J/molK×600K

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :