1.1 What is Chemistry?

17.4. Hess’s Law http://www.ck12.org

reverse the second known equation, NO 2 (g) will be a product. When reversing the direction of a chemical reaction,
the magnitude of the enthalpy change stays the same, but the sign is reversed.

NO(g)+^12 O 2 (g)→NO 2 (g) ∆H◦=−56 kJ/mol

Now we can add the two reactions:

1

2

N 2 (g)+

1

2

O 2 (g)→NO(g) ∆H◦f=90 kJ/mol

NO(g)+

1

2

O 2 (g)→NO 2 (g) ∆H◦=−56 kJ/mol

1

2

N 2 (g)+O 2 (g)+NO(g)→NO(g)+NO 2 (g) ∆H◦=34 kJ/mol

The∆H value for this transformation is the sum of the two simpler reactions. The final equation can also be
simplified. The NO cancels out, because it is on both sides of the equation, leaving only the desired transformation:
1
2 N^2 (g)+O^2 (g)→NO^2 (g) ∆H

◦

f=34 kJ/mol

Note that the enthalpy change for any transformation can be determined in this way; this method can be used for
more than just finding heat of formation values.

Example 17.8

What is∆Hrxnfor the following reaction?

CS 2 (l)+3O 2 (g)→CO 2 (g)+2SO 2 (g)

The following formation reactions have known∆H values:

C(s)+2S(s)→CS 2 (l) ∆Hf= 87 .9 kJ/mol

C(s)+O 2 (g)→CO 2 (g) ∆Hf=− 393 .5 kJ/mol

S(s)+O 2 (g)→SO 2 (g) ∆Hf=− 296 .8 kJ/mol

Answer:

Which components of the first known equation are part of the desired reaction? Carbon and sulfur are not involved
in the desired transformation, but one molecule of CS 2 is present as a reactant. If we reverse the first equation, we
get the following:

CS 2 (l)→C(s)+2S(s) ∆H=− 87 .9 kJ/mol

Now, look at the second equation. A single molecule of CO 2 is present as a product, which is what we want for our
final equation. Since CO 2 does not show up in any of the other equations, we can assume that this equation will be
used as is:

C(s)+O 2 (g)→CO 2 (g) ∆H=− 393 .5 kJ/mol

Finally, look at the third known equation. It has a single molecule of SO 2 as a product. However, we need two
molecules of SO 2 on the product side, so we multiply the entire equation, including its∆H value, by a factor of two:

2S(s)+2O 2 (g)→2SO 2 (g) ∆H=− 593 .6 kJ/mol

Now, add these three equations together: