CK-12-Chemistry Intermediate

http://www.ck12.org Chapter 17. Thermochemistry

reaction of solid carbon (graphite) with hydrogen gas. However, this is not an especially easy reaction for which
to measure the total enthalpy change. In contrast, enthalpy changes for combustion reactions are relatively easy to
measure. Theheat of combustionis the heat released when one mole of a substance is completely reacted with
oxygen gas.

FIGURE 17.10

The heats of combustion for carbon, hydrogen, and acetylene are shown below along with the balanced equation for
each process.

1. C(s,graphite)+O 2 (g)→CO 2 (g) ∆H=− 393 .5 kJ


2. H 2 (g)+

1

2

O 2 (g)→H 2 O(l) ∆H=− 285 .8 kJ

3. C 2 H 2 (g)+

5

2

O 2 (g)→2CO 2 (g)+H 2 O(l) ∆H=− 1301 .1 kJ

The combustion reactions are written with fractional coefficients for O 2 because the heats of combustion that are
found in a table are for the combustion of 1 mol of the given substance. To use Hess’s law, we need to determine
how the three equations above can be manipulated so that they can be added together to result in the desired equation
(the formation of acetylene from carbon and hydrogen).

In order to do this, we will go through the desired equation, one substance at a time—choosing the combustion
reaction from the equations numbered 1-3 above that contains that substance. It may be necessary to either reverse a
combustion reaction or multiply it by some factor in order to make it “fit” to the desired equation. The first reactant
is carbon, and the in the equation for the desired reaction, the coefficient of the carbon is a 2. We will therefore start
by writing the first combustion reaction with all of its coefficients doubled. Because we are doubling the coefficients,
we also need to double the value of∆H, since we are now looking at the heat released when two moles of graphite
are combusted instead of just one.

2C(s,graphite)+2O 2 (g)→2CO 2 (g) ∆H= 2 (− 393. 5 ) =− 787 .0 kJ

The second reactant is hydrogen, and its coefficient is a 1, as it is in the second combustion reaction. Therefore, that
reaction will be used as written.

H 2 (g)+

1

2

O 2 (g)→H 2 O(l) ∆H=− 285 .8 kJ

The product of the reaction is C 2 H 2 , and its coefficient is also a 1. In combustion reaction #3, the acetylene is a
reactant. Therefore, we will reverse reaction 3. When the reactants and products are reversed,∆H for the resulting
equation has the same numerical value but the opposite sign.