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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 that 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.)

Short Answer

Expert verified
The enthalpy change is +23.06 kJ/mol.

Step by step solution

01

Calculate the heat absorbed by the solution

Determine the heat absorbed by the solution using the formula:\[ q = mc\Delta T \]where \( q \) is the heat absorbed, \( m \) is the mass of the solution, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Inserting the given values:- \( m = 155.4 \) g (mass of the solution),- \( c = 4.2 \) \( \text{J/g}\cdot \text{K} \),- \( \Delta T = 16.2 - 18.6 = -2.4 \) °C (since each degree change in Celsius is equal to a change in Kelvin).Calculate:\[ q = 155.4 \times 4.2 \times (-2.4) = -1565.856 \text{ J} \]The negative sign indicates that heat is absorbed (endothermic process).
02

Convert heat from Joules to kilojoules

Convert the heat absorbed by the solution from Joules to kilojoules:\[ q = -1565.856 \text{ J} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = -1.565856 \text{ kJ} \]
03

Calculate the moles of NH4NO3

Find the number of moles of \( \text{NH}_4\text{NO}_3 \) using its molar mass. The molar mass of \( \text{NH}_4\text{NO}_3 \) is approximately \( 80.04 \text{ g/mol} \).Calculate the moles:\[ \text{moles of } \text{NH}_4\text{NO}_3 = \frac{5.44 \text{ g}}{80.04 \text{ g/mol}} = 0.0679 \text{ mol} \]
04

Calculate the enthalpy change per mole

Calculate the enthalpy change for dissolving \( \text{NH}_4\text{NO}_3 \) in water per mole:\[ \Delta H = \frac{-1.565856 \text{ kJ}}{0.0679 \text{ mol}} \approx -23.06 \text{ kJ/mol} \]Since the process is endothermic, the enthalpy change should be positive:\[ \Delta H = +23.06 \text{ kJ/mol} \]

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Key Concepts

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

Calorimetry
Calorimetry is a fascinating and practical method used to measure the heat exchange in chemical reactions or physical processes. In our everyday chemistry lab, a simple calorimeter such as a coffee-cup calorimeter gives us great insight into how energy changes occur.
Imagine having a cup that can hold liquid, and inside that liquid, a reaction occurs. The calorimeter's job is to measure how much heat is gained or lost by the solution.
In the given problem, the calorimeter measures the temperature change when ammonium nitrate (\(\text{NH}_4\text{NO}_3\)) dissolves in water. This temperature change allows us to calculate the heat involved in the dissolving process, thus providing data on the enthalpy change.
Endothermic Process
An endothermic process is a fascinating type of reaction where the reaction absorbs heat from its surroundings.
When you see the temperature of the surrounding getting cooler, an endothermic process is likely at play.
In the exercise, when ammonium nitrate is added to water, the temperature drops from \(18.6^{\circ}\text{C}\) to \(16.2^{\circ}\text{C}\).
This decrease in temperature suggests that the process absorbs heat from the water, making it cooler.
Therefore, dissolving ammonium nitrate in water is identified as an endothermic process, further aligning with how cold packs work!
  • The surrounding temperature drops.
  • Heat is absorbed by the reaction.
  • The solution feels colder to touch.
Specific Heat Capacity
Specific heat capacity is an essential concept that plays a vital role in understanding how substances absorb and release heat. It tells us how much heat is needed to raise 1 gram of a substance by 1 degree Celsius (or Kelvin).
Different substances have different capacities to hold and transfer heat. Water, for instance, has a specific heat capacity of \(4.2 \text{ J/g}\cdot \text{K}\), and that value is crucial in calculations.
In the coffee-cup calorimeter setup, we used this specific heat capacity to determine how much energy the water and the dissolved ammonium nitrate absorbed or released during the reaction process.
Thus, by using the formula \(q = mc\Delta T\), we could find the amount of heat absorbed by the solution.
Heat Absorption
Heat absorption occurs when energy from the surroundings enters a system or substance. In chemical terms, it's the amount of heat taken in during a reaction.
For the reaction involving \(\text{NH}_4\text{NO}_3\), heat absorption is determined by the change in temperature and the specific heat capacity of the solution.
When we calculated \(q = mc\Delta T\), we found the heat absorbed was negative \(-1565.856 \text{ J}\). The negative value signifies an endothermic process, where energy is absorbed from the surroundings.
When converted to \(\text{kJ}\), it helped us further deduce the enthalpy change for the reaction. Recall that energy is necessary for breaking bonds during dissolving, which is why heat is absorbed in the endothermic process.
This concept helps explain why substances like \(\text{NH}_4\text{NO}_3\) are used in products that require rapid cooling, such as cold packs.

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Most popular questions from this chapter

Three 45 -g ice cubes at \(0^{\circ} \mathrm{C}\) are dropped into \(5.00 \times\) \(10^{2} \mathrm{mL}\) of tea to make ice tea. The tea was initially at \(20.0^{\circ} \mathrm{C} ;\) when thermal equilibrium was reached, the final temperature was \(0^{\circ} \mathrm{C}\). How much of the ice melted and how much remained floating in the beverage? Assume the specific heat capacity of tea is the same as that of pure water.

The first law of thermodynamics is often described as another way of stating the law of conservation of energy. Discuss whether this is an accurate portrayal.

For each of the following, tell whether the process is exothermic or endothermic. (No calculations are required.) (a) \(\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) (b) \(2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{H}_{2} \mathrm{O}\left(\ell, 25^{\circ} \mathrm{C}\right) \longrightarrow \mathrm{H}_{2} \mathrm{O}\left(\ell, 15^{\circ} \mathrm{C}\right)\) (d) \(\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

A piece of chromium metal with a mass of \(24.26 \mathrm{g}\) is heated in boiling water to \(98.3^{\circ} \mathrm{C}\) and then dropped into a coffee-cup calorimeter containing \(82.3 \mathrm{g}\) of water at 23.3 "C. When thermal equilibrium is reached, the final temperature is \(25.6^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of chromium.

Sulfur \((2.56 \mathrm{g})\) is burned in a constant volume calorimeter with excess \(\mathrm{O}_{2}(\mathrm{g}) .\) The temperature increases from \(21.25^{\circ} \mathrm{C}\) to \(26.72^{\circ} \mathrm{C} .\) The bomb has a heat capacity of \(923 \mathrm{J} / \mathrm{K},\) and the calorimeter contains \(815 \mathrm{g}\) of water. Calculate the heat evolved, per mole of \(\mathrm{SO}_{2}\) formed, for the reaction $$\mathrm{S}_{8}(\mathrm{s})+8 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 8 \mathrm{SO}_{2}(\mathrm{g})$$

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