17.3. Enthalpy and Phase Transitions http://www.ck12.org
Some materials have negative heats of solution; the dissolution of one of these solutes in water is an exothermic
process. Heat is released, causing a net increase in the temperature of the solution. Conversely, other substances
have positive heats of solution. For example, the dissolution of potassium nitrate in water is an endothermic process.
The resulting absorption of energy causes the solution to become colder. Calculations involving heats of solutions
follow the same basic approach that we have used with other enthalpy problems.
Example 17.5
How much heat would be absorbed or released if we completely dissolved 0.45 moles of sulfuric acid in water?
Answer:
Since the heat of solution for H 2 SO 4 is negative, heat will be released during this process. The exact amount can be
calculated as follows:
∆H= 0 .45 mol H 2 SO 4 × 1 mol H−^96.^2 kJ 2 SO 4 =−43 kJ If we wanted to know the temperature change that this would
cause, we would need more information.
Example 17.6
If we dissolved 0.45 moles of sulfuric acid in 1.0 liter of water at 25.0°C, what would be the final temperature of the
solution, assuming no heat is lost to the surroundings? Assume that the specific heat of the solution is the same as
the specific heat of pure water.
Answer:
To answer this question, we will need to make use of the following equation:
∆H=m×c×∆T
Solving for the unknown variable (∆T), we get the following:
∆T=m∆×Hc
We already know that 43 kJ of energy will be released into the solution. Because the specific heat is in units of
joules, not kilojoules, we will want to use the value 43,000 J instead so that the units cancel out. The specific heat
of the solution is assumed to be the same as that of pure water (4.18 J/g•°C). Now, we need to find the mass of the
solution. At 25°C, 1.0 L of water has a mass of 1.0 kg (1,000 g). The mass of the sulfuric acid can be calculated as
follows:
0 .45 mol H 2 SO 4 ×^981 .mol H^08 g H 22 SOSO 44 =44 g H 2 SO 4
Therefore, the total mass of the solution would be 1,044 grams. Plugging these values into the above equation, we
get the following temperature change:
∆T= 1 , 04443 ,g^000 × 4 J. 18 J
g·◦C
= 9. 9 ◦C
The final temperature of the solution would be 34.9°C, which is 9.9°C higher than the initial temperature of 25.0°C.
These principles are applied when we use heat packs or cold packs for sore muscles. A heat pack contains water and
a solid such as calcium chloride or magnesium sulfate. When the pack is activated, the two materials mix and heat
is released. This is because the heats of solution are negative indicating an exothermic reaction. Some packs can get
as warm as 90°C. Cold packs use materials such as ammonium nitrate, which has a large positive heat of solution.
The endothermic mixing process can cool the solution down to just a few degrees Celsius.
Lesson Summary
- Specific heat and heat capacity provide information about enthalpy changes when the temperature of a sub-
stance is altered but no changes of state take place.