Question

Use standard heats of formation data from your textbook to calculate deltaH in kJ for the following reactions:

a). 2H2O2(l) --> 2H2O(l) + O2(g)

b). HCl(g) + NaOH(s) --> NaCl(s) + H2O(l)

Answer 1

a)

Given:
Hof(H2O2(l)) = -187.78 KJ/mol
Hof(H2O(l)) = -285.83 KJ/mol
Hof(O2(g)) = 0.0 KJ/mol

Balanced chemical equation is:
2 H2O2(l) ---> 2 H2O(l) + O2(g)

ΔHo rxn = 2*Hof(H2O(l)) + 1*Hof(O2(g)) - 2*Hof( H2O2(l))
ΔHo rxn = 2*Hof(H2O(l)) + 1*Hof(O2(g)) - 2*Hof( H2O2(l))
ΔHo rxn = 2*(-285.83) + 1*(0.0) - 2*(-187.78)
ΔHo rxn = -196.1 KJ/mol

Answer: -196.1 KJ/mol

b)

Given:
Hof(HCl(g)) = -92.307 KJ/mol
Hof(NaOH(s)) = -425.609 KJ/mol
Hof(NaCl(s)) = -411.153 KJ/mol
Hof(H2O(l)) = -285.83 KJ/mol

Balanced chemical equation is:
HCl(g) + NaOH(s) ---> NaCl(s) + H2O(l)

ΔHo rxn = 1*Hof(NaCl(s)) + 1*Hof(H2O(l)) - 1*Hof( HCl(g)) - 1*Hof(NaOH(s))
ΔHo rxn = 1*(-411.153) + 1*(-285.83) - 1*(-92.307) - 1*(-425.609)
ΔHo rxn = -179.067 KJ/mol

Answer: -179.067 KJ/mol