# 0.12 Measuring energy changes in chemical reactions  (Page 8/8)

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However, we do get a clearer understanding of why adding the reaction energies together gives the total energy of the overall reaction. Hess’ Law is a consequence of the Law of Conservation of Energy.

## Using hess’ law to measure reaction energy

It may be hard to remember now, but we started out with observations leading to Hess’ Law because we wanted to find a way to measure the energy of a reaction which can’t easily be found using calorimetry. Some reactions require special conditions which are hard to create in a laboratory where we can make measurements. We gave as an example fermentation of glucose to lactic acid:

${\text{2}}_{\text{2}}{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{→ 2 CH}}_{\text{3}}\text{CHOHCOOH}$

As we noted above, we can’t simply put glucose in a beaker and wait for it to turn into lactic acid and measure a temperature change of a water bath. The reaction just doesn’t happen without the assistance of enzymes in a cell.

So let’s use Hess’ Law, since we know that the energy of [link] will be the same as the sum of the energies of any set of reactions which adds up to [link] . We just need to pick some reactions which are easy to carry out in the laboratory so that we can measure the energies of these reactions.

The easiest reactions to conduct, particularly with molecules containing carbon, hydrogen, and oxygen, are almost always combustion reactions. We can pretty easily burn these compounds, reacting them with oxygen to form CO 2 (g) and H 2 O(g). We can also pretty easily measure the energies of these combustion reactions using calorimetry, just like before.

Here is what the experiments give us:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{(s) + 6O}}_{\text{2}}{\text{(g) → 6CO}}_{\text{2}}{\text{(g) + 6H}}_{\text{2}}\text{O(g)}$
${\text{2 CH}}_{\text{3}}{\text{CHOHCOOH + 6O}}_{\text{2}}{\text{(g) → 6CO}}_{\text{2}}{\text{(g) + 6H}}_{\text{2}}\text{O(g)}$

We can now follow a two step process which is equivalent to converting one glucose molecule into two lactic acid molecules. First, we burn the glucose, and the energy evolved is -2808 kJ. Second, hypothetically, we convert the CO 2 (g) and H 2 O(g) into lactic acid and oxygen. Although we can’t really do that hypothetical reaction, we don’t need to. The energy of that second step is just the negative of the energy we measure for the combustion of two moles of lactic acid, which is -2668 kJ. Using Hess’ Law, the overall energy for converting glucose to lactic acid then is just the measured energy of [link] plus the negative of the measured energy of [link] . This is equal to -140 kJ. We now have a means to measure the energy of a reaction which we can actually carry out!

This is a fairly general approach, applicable to most materials. By measuring the energies of combustion reactions and then summing those combustion reactions, we can calculate the energy of an overall reaction.

## Review and discussion questions

1. How can the temperature of water be elevated by doing work on it? Devise a way to measure the amount work required to raise the temperature of a sample of water by 1°C.
2. Assume you have two samples of two different metals, X and Z. The samples are exactly the same mass.

(a) Both samples are heated to the same temperature. Then each sample is placed into separate glasses containing identical quantities of cold water, initially at identical temperatures below that of the metals. The final temperature of the water containing metal X is greater than the final temperature of the water containing metal Z. Which of the two metals has the larger heat capacity? Explain your conclusion.

(b) If each sample, initially at the same temperature, is heated with exactly 100 J of energy, which sample has the higher final temperature?

3. Using data from [link] , provide two reasons with explanation why a hot object is much more efficiently cooled by placing it in water than leaving it in the open air, even when the air and the water are at the same temperature initially.
4. Explain how Hess' Law is a consequence of conservation of energy.
5. The enthalpy of formation of sucrose C 6 H 12 O 6 cannot be measured by the direct reaction of carbon, hydrogen and oxyben. Devise a method to measure ∆H f for sucrose. What would you measure and how do these measured quantities relate to the ∆H f for sucrose?

I only see partial conversation and what's the question here!
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research.net
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