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(FINAL) Chemistry Notebook 2016-17

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Step Four: liquid water boils<br />

Now, we continue to add energy and the water begins to<br />

boil.<br />

However, the temperature DOES NOT CHANGE. It<br />

remains at 100 during the time the water boils.<br />

Each mole of water will require a constant amount of<br />

energy to boil. That amount is named the molar heat of<br />

vaporization and its symbol is ΔH. The molar heat of<br />

vaporization is the energy required to boil one mole of a<br />

substance at its normal boiling point. One mole of liquid water, one mole of liquid benzene, one mole<br />

of liquid lead. It does not matter. Each substance has its own value.<br />

During this time, the energy is being used to overcome water molecules' attraction for each other,<br />

allowing them to move from close together (liquid) to quite far apart (the gas state).<br />

The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion<br />

between calories and Joules is 4.184 J = 1.000 cal.<br />

Typically, the term "heat of vaporization" is used with the "per gram" value.<br />

72.0 grams of liquid water is at 100.0 °C. It is going to boil AND stay at 100 degrees. This is an<br />

important point. While the water boils, its temperature will remain the same. We need to calculate the<br />

energy needed to do this.<br />

This summarizes the information needed:<br />

ΔH = 40.7 kJ/mol<br />

The mass = 72.0 g<br />

The molar mass of H2O = 18.0 gram/mol<br />

The calculation needed, using words & symbols is:<br />

q = (moles of water) (ΔH)<br />

We can rewrite the moles of water portion and make the equation like this:<br />

q = (grams water / molar mass of water) (ΔH)<br />

Why is this equation the way it is?<br />

Think about one mole of liquid water. That amount of water (one mole or 18.0 grams) needs 40.7<br />

kilojoules of energy to boil. Each mole of liquid water needs 40.7 kilojoules to boil. So the (grams<br />

water / molar mass of water) in the above equation calculates the amount of moles.<br />

With the numbers in place, we have:<br />

q = (72.0 g / 18.0 g mol¯1 ) (40.7 kJ / mol)<br />

So we calculate and get 162.8 kJ. We won't bother to round off right now since there is one more<br />

calculation to go. We're doing the fourth step now. When all five are done, we'll sum them all up.<br />

138

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