Hess's Law and Enthalpy

Question

Given:

CH3COOH(l) + 2O2(g) --> 2CO2(g) + 2H2O(l) =-871kJ/mole

C(graphite) + O2(g) --> CO2(g) =-394kJ/mole

H2(g) + 1/2O2(g) --> H2O(l) -286kJ/mole

Calculate the enthalpy for the reaction:

2C(graphite) + 2H2(g) +O2(g) --> CH3COOH(l)

Answer 1

Write the second equation and multiply all the coefficients and enthalpy by 2 to get

2C(graphite) +2O2 ===> 2CO2   - 788 kj/mole

 

Write the first equation in reverse and reverse the sign of the enthalpy.

2CO2 + 2H2O ==> CH3COOH + 2O2 +871 kj/mole

Write the third equation as given except multiply by 2,

2H2 + O2 ==> 2 H2O - 572 kj/mole

Now, add the 3 modified equations and cancel out similar terms on each side of the resulting equation. Algebraically add the enthalpies as well.

2C(graphite) +2O2 ===> 2CO2   - 788 kj/mole

2CO2 + 2H2O ==> CH3COOH + 2O2 +871 kj/mole

2H2 + O2 ==> 2 H2O - 572 kj/mole

 

2C (graphite) + 2H2(g) + O2(g) ===> CH#COOH(l) ( -788 +  +871 -572 = -489 kj/mole)

Please check all math steps.