dHrxn = (n*dHf products( - (n*dHf reactants)
2923.0 kJ = (4*dHCO2 + 6*dHH2O) - (2*dHC2H6). Look up dH H2O and dH C2H6, substitute and solve for dH CO2.
A scientist measures the standard enthalpy change for the following reaction to be -2923.0 kJ :
2C2H6(g) + 7 O2(g) 4CO2(g) + 6 H2O(g)
Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CO2(g) is _________
2 answers
Horxn = (4 mol)Hf(CO2(g)) + (6 mol)Hf(H2O(g)) - [(2 mol)Hf(C2H6(g)) + (7 mol)Hf(O2(g))] = -2909.8 kJ
Solve for Hf(CO2(g)) in the above equation:
Hf(CO2(g)) = [Horxn - (6 mol)Hf(H2O(g)) + (2 mol)Hf(C2H6(g)) + (7 mol)Hf(O2(g))] / (4 mol)
= [(-2909.8 kJ) - (6 mol)(-241.8 kJ/mol) + (2 mol)(-84.7 kJ/mol) + (7 mol)(0.0 kJ/mol)]/(4 mol)
=-407.1 kJ/mol
Solve for Hf(CO2(g)) in the above equation:
Hf(CO2(g)) = [Horxn - (6 mol)Hf(H2O(g)) + (2 mol)Hf(C2H6(g)) + (7 mol)Hf(O2(g))] / (4 mol)
= [(-2909.8 kJ) - (6 mol)(-241.8 kJ/mol) + (2 mol)(-84.7 kJ/mol) + (7 mol)(0.0 kJ/mol)]/(4 mol)
=-407.1 kJ/mol
1 answer