/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 (a) When a 4.25 -g sample of sol... [FREE SOLUTION] | 91影视

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(a) When a 4.25 -g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter (Figure 5.18), the temperature drops from 22.0 to \(16.9^{\circ} \mathrm{C}\) . Calculate \(\Delta H\left(\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) for the solution process: $$\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)$$ Assume that the specific heat of the solution is the same as that of pure water. (b) Is this process endothermic or exothermic?

Short Answer

Expert verified
The enthalpy change for the dissolution of NH鈧凬O鈧 is \(-24.065\,\text{kJ/mol}\), and the process is exothermic.

Step by step solution

01

List the given information

- Mass of NH鈧凬O鈧 (m鈧) = 4.25g - Mass of H鈧侽 (m鈧) = 60.0g - Initial temperature (T鈧) = 22.0掳C - Final temperature (T鈧) = 16.9掳C - Specific heat capacity of water (c) = 4.18 J/(g掳C) Note: The specific heat capacity of the solution is assumed to be the same as that of pure water.
02

Calculate the temperature change (鈭員)

We will first calculate the temperature change by subtracting the final temperature from the initial temperature. 鈭員 = T鈧 - T鈧 = 16.9掳C - 22.0掳C = -5.1掳C
03

Calculate the energy change (q)

Now, we will determine the energy change (q) using the equation q = mc鈭員. Since the specific heat capacity of the solution is the same as that of pure water, we can use the mass of water (m鈧) in our calculations. q = m鈧 脳 c 脳 鈭員 = 60.0g 脳 4.18 J/(g掳C) 脳 (-5.1掳C) q = -1277.64 J
04

Convert the mass of NH鈧凬O鈧 to moles

To find the moles of NH鈧凬O鈧, we will divide the mass (m鈧) by the molar mass. Molar mass of NH鈧凬O鈧 = (1 * 14.01 + 4 * 1.01) + (1 * 14.01 + 3 * 16.00) = 18.05 + 62.01 = 80.06 g/mol moles of NH鈧凬O鈧 = m鈧 / (molar mass) = 4.25g / 80.06 g/mol moles of NH鈧凬O鈧 = 0.0531 mol
05

Calculate 鈭咹

Now, we will find 鈭咹 by dividing the energy change (q) by the moles of NH鈧凬O鈧 and converting the result to kJ/mol. 鈭咹 = (q) / (moles of NH鈧凬O鈧) = -1277.64 J / 0.0531 mol 鈭咹 = -24,065 J/mol Since we need to convert the result to kJ/mol, divide by 1000: 鈭咹 = -24.065 kJ/mol
06

Determine if the process is endothermic or exothermic

The value of 鈭咹 is negative, which means the process is exothermic. This indicates that heat is released during the dissolution of NH鈧凬O鈧. In conclusion, the enthalpy change for the dissolution of NH鈧凬O鈧 is -24.065 kJ/mol, and the process is exothermic.

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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 an essential tool in chemistry used to measure the heat changes during a reaction or process. Made from a simple coffee cup with a lid, it involves placing a sample in a liquid鈥攐ften water鈥攖o observe the temperature change. This change helps determine the heat absorbed or released during a chemical reaction.

The setup takes advantage of the insulating properties of a coffee cup to minimize heat exchange with the environment. This way, we can assume that most of the heat transfer occurs between the solution and the substances inside the calorimeter.

A coffee-cup calorimeter operates under constant pressure conditions, which is ideal for studying processes that occur in solution, such as dissolving a solute like ammonium nitrate. By knowing the mass of the substances involved, the specific heat capacity, and the change in temperature, it allows us to calculate the energy change of the process.
Specific Heat Capacity
Specific heat capacity is a property of a material that defines how much heat energy it needs to change its temperature. For water, this value is 4.18 J/(g掳C), meaning that it takes 4.18 joules of energy to raise the temperature of one gram of water by one degree Celsius.

In the context of our problem, we assume the solution's specific heat capacity is the same as water's. This simplification helps to calculate the heat gained or lost during the reaction easily. The equation used is:\[ q = m \times c \times \Delta T \]Where:
  • \( q \) is the heat absorbed or released (in Joules).
  • \( m \) is the mass of the solution (in grams).
  • \( c \) is the specific heat capacity.
  • \( \Delta T \) is the temperature change.
Despite being a simple concept, understanding specific heat capacity is vital as it allows us to determine how different substances respond to heat and how they affect temperature in reactions.
Endothermic and Exothermic Processes
In chemistry, reactions are typically categorized into endothermic and exothermic based on their heat exchange with the environment.

**Endothermic processes** absorb heat from their surroundings. These processes feel cold to the touch because they require energy. They have a positive enthalpy change (\(\Delta H\)). This means energy is taken in, essential when bonds are being broken or new product phases are being formed.

**Exothermic processes**, on the other hand, release heat into the surroundings. This release can often be felt as warmth. They have a negative enthalpy change (\(\Delta H\)), as seen in our problem where ammonium nitrate dissolves in water. This indicates that the process releases energy, generally because new bonds are being formed that are more stable.

Identifying whether a process is endothermic or exothermic is fundamental in understanding chemical reactions' energy dynamics and predicting their behavior under different conditions.

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Most popular questions from this chapter

(a) Which releases the most energy when metabolized, 1 \(\mathrm{g}\) of carbohydrates or 1 \(\mathrm{g}\) of fat? (b) A particular chip snack food is composed of 12\(\%\) protein, 14\(\%\) fat, and the rest carbohydrate. What percentage of the calorie content of this food is fat? (c) How many grams of protein provide the same fuel value as 25 of fat?

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) is blended with gasoline as an automobile fuel. (a) Write a balanced equation for the combustion of liquid ethanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming \(\mathrm{H}_{2} \mathrm{O}(g)\) as a product. (c) Calculate the heat produced per liter of ethanol by combustion of ethanol under constant pressure. Ethanol has a density of 0.789 \(\mathrm{g} / \mathrm{mL}\) (d) Calculate the mass of \(\mathrm{CO}_{2}\) produced per ky of heat emitted.

A sample of a hydrocarbon is combusted completely in \(\mathrm{O}_{2}(g)\) to produce \(21.83 \mathrm{g} \mathrm{CO}_{2}(g), 4.47 \mathrm{g} \mathrm{H}_{2} \mathrm{O}(g),\) and 311 \(\mathrm{kJ}\) of heat. (a) What is the mass of the hydrocarbon sample that was combusted? (b) What is the empirical formula of the hydrocarbon? (c) Calculate the value of \(\Delta H_{f}^{\circ}\) per empirical-formula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.

Diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l),\) a flammable compound that was once used as a surgical anesthetic, has the structure $$\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3}$$ The complete combustion of 1 mol of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) yields \(\Delta H^{\circ}=-2723.7 \mathrm{kJ}\) . (a) Write a balanced equation for the combustion of 1 \(\mathrm{mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l) .\) (b) By using the information in this problem and data in Table \(5.3,\) calculate \(\Delta H_{f}^{\circ}\) for diethyl ether.

Under constant-volume conditions, the heat of combustion of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) is 15.57 \(\mathrm{kJ} / \mathrm{g}\) . A 3.500 -g sample of glucose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to \(24.72^{\circ} \mathrm{C}\) (a) What is the total heat capacity of the calorimeter? (b) If the size of the glucose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

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