Hess’s Law

The overall ∆H for a reaction will be the same as the sum of the individual steps taken in the process. For example: Let’s say we need to know the ∆H for the following reaction: C(s) + 2H2(g)  CH4(g) Let’s assume that this reaction is difficult or impossible to actually measure the ∆H for. Then assume that we do know the following information for some other reactions: C(s) + O2(g)  CO2(g) ∆H= -393.5 kJ/mol H2(g) + ½ O2(g)  H2O(l) ∆H= -285.8 kJ/mol CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ∆H= -890.8 kJ/mol Because these three contain the components of our original equation (in blue above), we might be able to manipulate them algebraically to add up to the desired equation.

Desired equation: C(s) + 2H2(g)  CH4(g) ∆H = Desired equation: C(s) + 2H2(g)  CH4(g) ∆H = ??? Go through the 3 given equations and find each piece of the desired equation. For example, carbon is found in the very first equation: C(s) + O2(g)  CO2(g) ∆H= -393.5 kJ/mol Hydrogen is found in the second equation, however the second equation only has 1 mole of hydrogen. Therefore we must multiply the equation by 2, including the heat. 2(H2(g) + ½ O2(g)  H2O(l) ∆H= -285.8 kJ/mol) Rewritten: 2H2(g) + O2(g)  2H2O(l) ∆H= -571.56 kJ/mol The CH4 is found in the third equation, but is on the reactant-side. We need it on the product side, so we flip the equation, but also flip the sign of the heat. CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ∆H= -890.8 kJ/mol Becomes: CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) ∆H= +890.8 kJ/mol

The last step is to add up the 3 equations, cancelling any substances that show up on either side of the arrow. We’ll also need to add up the heats: C(s) + O2(g)  CO2(g) ∆H= -393.5 kJ/mol 2H2(g) + O2(g)  2H2O(l) ∆H= -571.56 kJ/mol CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) ∆H= +890.8 kJ/mol C(s) + 2H2(g)  CH4(g) ∆H = -74.3 kJ/mol

Try this one on your own. Determine the heat of reaction for the reaction for the following: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ (hint: Because O2 shows up in two different equations, just skip it. If you focus on the other three substances, the 5O2 should work out naturally on the left.) Answer: H = -906.3 kJ