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Chapter 5: Problem 33

Adding \(5.44 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s})\) to \(150.0 \mathrm{g}\) of water in a coffee-cup calorimeter (with stirring to dissolve the salt) resulted in a decrease in temperature from \(18.6^{\circ} \mathrm{C}\) to \(16.2^{\circ} \mathrm{C} .\) Calculate the enthalpy change for dissolving \(\mathrm{NH}_{4} \mathrm{NO}_{3}(\mathrm{s})\) in water, in \(\mathrm{kJ} / \mathrm{mol}\). Assume the solution (whose mass is \(155.4 \mathrm{g})\) has a specific heat capacity of \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\) (Cold packs take advantage of the fact that dissolving ammonium nitrate in water is an endothermic process.) (IMAGE CAN'T COPY)

Short Answer

Expert verified
The enthalpy change is \(23.03\, \mathrm{kJ/mol}\).

Step by step solution

01

Calculate the temperature change

Determine the change in temperature (螖T) of the solution. Use the formula: \[\Delta T = T_{\text{final}} - T_{\text{initial}}\]Given, \(T_{\text{initial}} = 18.6^{\circ} \mathrm{C}\) and \(T_{\text{final}} = 16.2^{\circ} \mathrm{C}\), so:\[\Delta T = 16.2 - 18.6 = -2.4^{\circ} \mathrm{C}\]The negative sign indicates a decrease in temperature.
02

Calculate the heat absorbed by the solution

Use the formula for heat absorbed or released: \[q = m \cdot c \cdot \Delta T\]where \(m\) is the mass of the solution, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change.\(m = 150.0 \mathrm{g(water)} + 5.44 \mathrm{g(salt)} = 155.44 \mathrm{g}\),\(c = 4.2 \mathrm{J/g\cdot K}\), and \(\Delta T = -2.4\, ^{\circ} \mathrm{C}\). So,\[q = 155.44 \times 4.2 \times (-2.4) = -1565.47 \mathrm{J}\]The negative sign indicates heat absorption from the surroundings.
03

Convert the heat to kilojoules

Since energy needs to be calculated in kJ, convert the heat from J to kJ:\[q = -1565.47 \mathrm{J} = -1.56547 \mathrm{kJ}\]
04

Calculate moles of NH4NO3

Determine the number of moles of \(\mathrm{NH}_4\mathrm{NO}_3\). Use the formula:\[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\]Given, mass of \(\mathrm{NH}_4\mathrm{NO}_3 = 5.44\, \mathrm{g}\) and molar mass \(= 80.04\, \mathrm{g/mol}\), so:\[\text{moles of } \mathrm{NH}_4\mathrm{NO}_3 = \frac{5.44}{80.04} = 0.06796\, \text{mol}\]
05

Calculate enthalpy change per mole

Enthalpy change \(\Delta H\) for dissolving is calculated as:\[\Delta H = \frac{q}{\text{moles of } \mathrm{NH}_4\mathrm{NO}_3}\]Substitute the values:\[\Delta H = \frac{-1.56547}{0.06796} = -23.03\, \mathrm{kJ/mol}\]This result is adjusted for a positive value since the process is endothermic and absorbs heat, making \(\Delta H = 23.03\, \mathrm{kJ/mol}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coffee-Cup Calorimeter
A coffee-cup calorimeter is a simple device used in thermal experiments to measure the heat exchanged between the system and its surroundings. It's essentially two nested Styrofoam cups with a lid, often containing a thermometer and a stirrer. This setup provides excellent thermal insulation, trapping heat inside and reducing heat loss to the environment.

In experiments, solutions are typically mixed in this calorimeter, and the change in temperature is carefully observed. The purpose is to determine properties like enthalpy change during chemical reactions. The advantage of using a coffee-cup calorimeter is that it's relatively easy to use and cost-effective for conducting basic calorimetry experiments.
  • Made from Styrofoam for insulation
  • Used to measure temperature changes in solutions
  • Helps calculate energy changes in chemical reactions
Endothermic Process
When ammonium nitrate ( ext{NH}_4 ext{NO}_3) dissolves in water, the reaction is endothermic. This means it absorbs heat from its surroundings, leading to a decrease in the temperature of the solution. Endothermic processes are characterized by a positive enthalpy change ( ext{螖H}).

In an endothermic reaction, such as this one, you can feel the solution getting colder to the touch. This is because the system absorbs energy in the form of heat from the water in the calorimeter, thereby decreasing the temperature.
  • Absorbs heat from surroundings
  • Characterized by a temperature drop
  • Positive enthalpy change
Specific Heat Capacity
Specific heat capacity is a property that defines how much heat energy is needed to raise the temperature of 1 gram of a substance by 1 degree Celsius (or 1 Kelvin). It's denoted by the symbol ext{c} and is expressed in units of ext{J/g鈥}.

In the exercise, water and the dissolved ammonium nitrate have a specific heat capacity of 4.2 ext{J/g鈥}. This means that for each gram of the solution, 4.2 Joules of energy are required to change the temperature by 1鈩�. This property helps calculate the heat absorbed or released by the solution in thermal processes.
  • Defines energy needed per unit mass for temperature change
  • Varies between different substances
  • Essential in calculating heat flow
Temperature Change
Temperature change ( ext{螖T}) is a crucial factor in calorimetry, representing the difference between the initial and final temperatures of a system. In our example, the temperature change was calculated as 饾憞_{ ext{final}} - 饾憞_{ ext{initial}}, which turned out to be -2.4鈩�. This negative value indicated that the temperature dropped during the process.

Understanding temperature change is vital because it allows us to determine the direction and magnitude of heat flow within a reaction. In endothermic reactions, as more heat is absorbed, the indicator (thermometer) reflects a falling temperature.
  • Difference between initial and final temperatures
  • Determines heat flow direction
  • Shows endothermic nature when negative

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Most popular questions from this chapter

An "ice calorimeter" can be used to determine the specific heat capacity of a metal. A piece of hot metal is dropped onto a weighed quantity of ice. The energy transferred from the metal to the ice can be determined from the amount of ice melted. Suppose you heated a 50.0 -g piece of silver to \(99.8^{\circ} \mathrm{C}\) and then dropped it onto ice. When the metal's temperature had dropped to \(0.0^{\circ} \mathrm{C},\) it is found that \(3.54 \mathrm{g}\) of ice had melted. What is the specific heat capacity of silver?

The specific heat capacity of copper metal is \(0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\) How much energy is required to heat \(168 \mathrm{g}\) of copper from \(-12.2^{\circ} \mathrm{C}\) to \(+25.6^{\circ} \mathrm{C} ?\)

Suppose you burned \(1.500 \mathrm{g}\) of benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H},\) in a constant volume calorimeter and found that the temperature increased from \(22.50^{\circ} \mathrm{C}\) to \(31.69^{\circ} \mathrm{C} .\) The calorimeter contained \(775 \mathrm{g}\) of water, and the bomb had a heat capacity of \(893 \mathrm{J} / \mathrm{K}\). Calculate \(\Delta U\) per mole of benzoic acid. (IMAGE CAN'T COPY)

Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) boils at \(78.29^{\circ} \mathrm{C} .\) How much energy, in joules, is required to raise the temperature of \(1.00 \mathrm{kg}\) of ethanol from \(20.0^{\circ} \mathrm{C}\) to the boiling point and then to change the liquid to vapor at that temperature? (The specific heat capacity of liquid ethanol is \(2.44 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\) and its enthalpy of vaporization is \(855 \mathrm{J} / \mathrm{g} .\) )

You wish to know the enthalpy change for the formation of liquid \(\mathrm{PCl}_{3}\) from the elements. $$ \mathrm{P}_{4}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{3}(\ell) \quad \Delta_{\mathrm{r}} H^{\circ}=? $$ The enthalpy change for the formation of \(\mathrm{PCl}_{5}\) from the elements can be determined experimentally, as can the enthalpy change for the reaction of \(\mathrm{PCl}_{3}(\ell)\) with more chlorine to give \(\mathrm{PCl}_{5}(\mathrm{s}):\) \(\begin{aligned} \mathrm{P}_{4}(\mathrm{s})+10 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{s}) & \\ \Delta_{r} H^{\circ} &=-1774.0 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \\\ \mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{PCl}_{5}(\mathrm{s}) & \\ \Delta_{\mathrm{r}} H^{\circ} &=-123.8 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \end{aligned}\) Use these data to calculate the enthalpy change for the formation of 1.00 mol of \(\mathrm{PCl}_{3}(\ell)\) from phosphorus and chlorine.
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