Hess’s Law
Name: ___________________________
Definition of Hess’s Law = If a series of reactions are added together, the enthalpy change
for the net reaction will be the sum of the enthalpy changes for the individual steps
Steps to solve a Hess’s Law problem:
1) Look at your overall reaction goal (AKA “net reaction”): find ways to match up
reactants and products of smaller equations to look the same
 FLIP = reverse the reactants and products; change sign of H
 MULTIPLE or DIVIDE by a factor = apply to the entire equation; do the same to H
2) Cancel out like terms on opposite sides of the arrow.
The only terms that should be left are those in the overall net reaction
3) Add H’s of smaller reactions to get H of the overall net reaction.
Practice:
1) Calculate ∆H for the reaction: H2CO + O2  H2CO3
∆Hnet = -67 kJ
H2CO3  H2O + CO2
∆H = +37 kJ
H2CO + O2  H2O + CO2
∆H = -30 kJ
2) Calculate the net enthalpy change for the following reaction:
N2 + O2  2NO
H = +180 kJ
2NO2  2NO + O2
H = +112 kJ
N2 + 2O2  2NO2
∆Hnet = +68 kJ
3)
overall:
Fe2O3 + 3CO  2Fe + 3CO2
H = -27 kJ
C + CO2  2CO
H = +172 kJ
2Fe2O3 + 3C  4 Fe + 3CO2
∆Hnet = +462 kJ
4) Calculate ∆H for the reaction 2F2(g)+ 2H2O(l)  4HF(g) + O2(g), given: ∆Hnet = -512.8 kJ
H2(g) + F2(g)  2HF(g)
∆H = -542.2 kJ
2H2(g) + O2(g)  2H2O(l)
∆H = -571.6 kJ
5)
H2S + 1½ O2  H2O + SO2
H = -563 kJ
CS2 + 3O2  CO2 + 2SO2
H = -1075 kJ
overall:
CS2 + 2H2O  CO2 + 2H2S
∆Hnet = +51 kJ
6) Calculate the net enthalpy change for the reaction Cu2O + ½ O2  2CuO , given:
Cu2O + H2O2  2CuO + H2O
∆H = -239 kJ
2H2O2  2H2O + O2
∆H = -196 kJ
∆Hnet = -141 kJ
7) From the following enthalpies of reaction, calculate ∆H for: 2CO2 + 3H2O  C2H6O + 3O2
2C2H4O + 2H2O  2C2H6O + O2
∆H = 407 kJ
C2H4O + 2½ O2  2CO2 + 2H2O
∆H = -1167 kJ
∆Hnet = +1370.5 kJ
8) Calculate the ∆H for the reaction C2H4 (g) + 6F2 (g)  2 CF4 (g) + 4 HF(g)
H2 + F2  2 HF
∆H = - 537kJ
C + 2 F2  CF4
∆H = - 680 kJ
2 C + 2 H2  C2H4
∆H = + 52.3 kJ
∆Hnet = -2486.3 kJ
9) Calculate H for the reaction 4 NH3 + 5 O2  4 NO + 6 H2O, given:
N2 + O2  2 NO
∆H = -180.5 kJ
N2 + 3 H2  2 NH3
∆H = -91.8 kJ
2 H2 + O2  2 H2O
∆H = -483.6 kJ
∆Hnet = -1628.2 kJ