Question

A scientist measures the standard enthalpy change for the following reaction to be -573.8 kJ: H.CO(g) + O2() CO() + H2O(1) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O() is mol Submit Answer Retry Entire Group 3 more group attempts remaining

Answer 1

Given:

ΔHo rxn = -573.8 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

Hof(CO2(g)) = -393.509 KJ/mol

Hof(H2O(l)) = -285.83 KJ/mol

Balanced chemical equation is:

H2CO(g) + O2(g) ---> CO2(g) + H2O(l)

ΔHo rxn = 1*Hof(CO2(g)) + 1*Hof(H2O(l)) - 1*Hof( H2CO(g)) - 1*Hof(O2(g))

-573.8 = 1*(-393.509) + 1*(-285.83) - 1*Hof(H2CO(g)) - 1*(0.0)

Hof(H2CO(g)) = -105.539 KJ/mol

Answer: -105.54 KJ/mol