Question

30) Given the following reactions and their enthalpy changes, calculate the enthalpy change for

2NO₂(g) ---> N₂O₄(g)

Equations	Change in Heat Energy
N2(g) + 2O2(g) ---> 2NO2 (g)	+67.8kJ
N2(g) + 2O2(g) ---> N2O4(g)	+9.67kJ

32) Calculate ΔH for the process Hg₂Cl₂(s) ---> 2Hg(l) + Cl₂(g) from the following information:

Hg(l) + Cl₂(g) --> HgCl₂(s) ΔH = - 224kJ

Hg(l) + HgCl₂(s) ---> Hg₂Cl₂(s) ΔH = - 41.2kJ

Answer 1

ΔH rxn = ΔH products - ΔH reactants

ΔH of elements at standard state is zero

30)

2NO2 (g) ---> N2O4 (g)

ΔH rxn = ΔH of N2O4 - ΔH of 2*NO2

From given table

67.8 kJ = ΔH of 2*NO2 - (0 + 2*0)

ΔH of NO2 = 67.8 kJ/2 => 33.9 kJ

9.67 kJ = ΔH of N2O4 - (0+ 2*0)

ΔH of N2O4 = 9.67 kJ

So ΔH rxn = ΔH of N2O4 - ΔH of 2*NO2

ΔH rxn = 9.67 - (2*33.9) kJ

ΔH rxn = - 58.13 kJ

32)

-224 kJ = ΔH of HgCl2(s) - ( 0+0)

ΔH of HgCl2 (s) = - 224 kJ

-41.2 kJ = ΔH of Hg2Cl2 (s) - (0 + -224)

ΔH of Hg2Cl2 (s) = -41.2 - 224 kJ => - 265.2 kJ

For the reaction Hg2Cl2 (s) ---> 2 Hg(l) + Cl2 (g)

ΔH rxn = (2*0 + 0) - (-265.2 kJ)

ΔH rxn = 265.2 kJ