The Foundations of Chemistry

(Marcin) #1
EXAMPLE 15-3 Thermochemical Equations
When 2.61 grams of dimethyl ether, CH 3 OCH 3 , is burned at constant pressure, 82.5 kJ of
heat is given off. Find Hfor the reaction

CH 3 OCH 3 ()3O 2 (g)88n2CO 2 (g)3H 2 O()
Plan
We scale the amount of heat given off in the experiment to correspond to the amount of
CH 3 OCH 3 shown in the balanced equation.
Solution



1450 kJ/mol rxn

Because heat is given off, we know that the reaction is exothermic and the value of His nega-
tive, so

H1450 kJ/mol rxn

You should now work Exercise 19.

EXAMPLE 15-4 Thermochemical Equations
Write the thermochemical equation for the reaction in Example 15-2.

Plan
We must determine how muchreaction occurred — that is, how many moles of reactants were
consumed. We first multiply the volume, in liters, of each solution by its concentration in
mol/L (molarity) to determine the number of moles of each reactant mixed. Then we identify
the limiting reactant. We scale the amount of heat released in the experiment to correspond
to the number of moles of that reactant shown in the balanced equation.
Solution
Using the data from Example 15-2,

__? mol CuSO 4 0.0500 L0.0200 mol CuSO 4

__? mol NaOH0.0500 L0.0300 mol NaOH

We determine which is the limiting reactant (see Section 3-3).

Required Ratio Available Ratio



More CuSO 4 is available than is required to react with the NaOH. Thus, 0.846 kJ of heat was
given off during the consumption of 0.0300 mol of NaOH. The amount of heat given off per
“mole of reaction” is



56.4 kJ given off

mol rxn

2 mol NaOH

mol rxn

0.846 kJ given off

0.0300 mol NaOH

__? kJ released

mol rxn

0.667 mol CuSO 4

1.00 mol NaOH

0.0200 mol CuSO 4

0.0300 mol NaOH

0.50 mol CuSO 4

1.00 mol NaOH

1 mol CuSO 4

2 mol NaOH

0.600 mol NaOH

1.00 L

0.400 mol CuSO 4

1.00 L

1 mol CH 3 OCH 3

mol rxn

46.0 g CH 3 OCH 3

mol CH 3 OCH 3

82.5 kJ given off

2.61 g CH 3 OCH 3

? kJ given off

mol rxn

NaOH is the limiting reactant.


600 CHAPTER 15: Chemical Thermodynamics

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