The use of a constant-pressure calorimeter is illustrated in Example \(\PageIndex{3}\).

Example \(\PageIndex{1}\)

When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm^{3}. What is Δ*H*_{soln} (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water.

**Given: **mass of substance, volume of solvent, and initial and final temperatures

**Asked for: **Δ*H*_{soln}

**Strategy:**

- Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution.
- Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into Equation 5.5.8.
- Use the molar mass of KOH to calculate Δ
*H*_{soln}.

**Solution:**

**A** To calculate Δ*H*_{soln}, we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is

\[ \left (100.0 \; mL\; H2O \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; KOH=104.72 \; g \]

The temperature change is (34.7°C − 23.0°C) = +11.7°C.

**B** Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus

\[ q_{calorimater}=mC_{s} \Delta T =\left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right )\left ( 11.7 \; ^{o}C \right )=5130 \; J =5.13 \; lJ \]

The temperature of the solution increased because heat was absorbed by the solution (*q* > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation \(\ref{5.5.8}\), we see that

Δ*H*_{rxn} = −*q*_{calorimeter} = −5.13 kJ

This experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the *release* of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic.

**C** The last step is to use the molar mass of KOH to calculate Δ*H*_{soln}—the heat released when dissolving 1 mol of KOH:

\[ \Delta H_{soln}= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right )=-57.2 \; kJ/mol \]

Exercise \(\PageIndex{1}\)

A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example \(\PageIndex{1}\), find Δ*H*_{soln} for NH_{4}Br (in kilojoules per mole).

**Answer**

16.6 kJ/mol