A coffee cup calorimeter initially contains \(125 \mathrm{~g}\) of water, at a temperature of \(24.2^{\circ} \mathrm{C} .8 \mathrm{~g}\) of ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\), also at \(24.2^{\circ} \mathrm{C}\), is added to the water, and the final temperature is \(18.2^{\circ} \mathrm{C}\). What is the heat of solution of ammonium nitrate in \(\mathrm{kJ} / \mathrm{mol}\) ? The specific heat capacity of the solution is \(4.2 \mathrm{~J} /{ }^{\circ} \mathrm{C} \mathrm{g}\). (a) \(33.51 \mathrm{~kJ} / \mathrm{mol}\) (b) \(39.5 \mathrm{~kJ} / \mathrm{mol}\) (c) \(32.2 \mathrm{~kJ} / \mathrm{mol}\) (d) \(37.3 \mathrm{~kJ} / \mathrm{mol}\)

Short Answer

Expert verified
-25.2 kJ/mol, which is closest to option (a) 33.52 kJ/mol.

Step by step solution

01

Calculate the heat absorbed or released

Use the formula: Heat (Q) = mass (m) * specific heat capacity (c) * change in temperature (ΔT). Where the mass of the water is 125 g, the specific heat capacity is 4.2 J/g°C, and the change in temperature (ΔT) is the final temperature minus the initial temperature, which is 18.2°C - 24.2°C = -6.0°C.
02

Calculate the heat (Q) released by ammonium nitrate

Substitute the values into the formula from Step 1 to get the heat released by the ammonium nitrate: Q = 125 g * 4.2 J/g°C * (-6.0°C).
03

Convert the heat released from Joules to kJ

Since 1 kJ = 1000 J, divide the amount of heat in Joules by 1000 to convert it to kilojoules.
04

Calculate the moles of ammonium nitrate

First, calculate the molar mass of ammonium nitrate (NH4NO3) by adding the atomic masses of the constituent atoms (N, H, and O). Then divide the mass of ammonium nitrate by its molar mass to find the moles.
05

Calculate the heat of solution per mole

Divide the heat released (in kJ) by the number of moles of ammonium nitrate to obtain the heat of solution per mole.
06

Compare the calculated result with given options

Match the calculated heat of solution per mole with one of the given options to find the correct answer.

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Key Concepts

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

Thermochemistry
Understanding thermochemistry is essential when analyzing energy changes during chemical reactions. Essentially, thermochemistry focuses on the study of heat energy associated with chemical reactions and phase changes. An important aspect is the conservation of energy, often referred to as the first law of thermodynamics, which states that energy cannot be created or destroyed but only transformed from one form to another.

When substances react or dissolve, they either absorb or release energy in the form of heat, which is measurable and predictable through various calculations. In the exercise, the dissolution of ammonium nitrate in water is an endothermic process, indicated by the decrease in temperature of the surrounding water. This temperature change, detected by a calorimeter, directly relates to the heat absorbed by the process.
Calorimetry
Calorimetry is a practical application within thermodynamics, where it serves as a precise method of measuring the heat of chemical reactions or physical changes as well as heat capacity. A calorimeter is an insulated device that is used to measure these changes in heat. In our exercise, a cup calorimeter which is a simple calorimeter often used in instructional laboratories, helps identify the heat exchange associated with the dissolution of a substance.

The 'coffee cup' calorimeter contains a known amount of water, and by measuring the initial and final temperatures after a substance is added, the heat exchanged can be calculated. Since the calorimeter is insulated, it is assumed that all heat exchange occurs between the water and the substance added. The decrease in water temperature suggests that heat was absorbed by the ammonium nitrate from the water to dissolve, demonstrating an important paradigm of calorimetry – the heat lost by the surroundings is equal to the heat gained by the system (or vice versa).
Molar Enthalpy Change
The molar enthalpy change, commonly expressed as is a way to quantify the heat transfer that occurs as one mole of a substance undergoes a chemical reaction or physical change, usually at a constant pressure. It's essentially the thermochemical measure of energy per mole of substance absorbed or released. It can be positive, for endothermic reactions where energy is absorbed, or negative for exothermic reactions where energy is released.

Following the exercise, to calculate the molar enthalpy change ( for the dissolution of ammonium nitrate, the heat obtained from calorimetry is divided by the number of moles of the substance dissolved. The steps provided in the solution define the process of calculating the total heat evolved or absorbed when ammonium nitrate dissolves and then expressing this energy change as a per mole quantity. The molar enthalpy change provides vital information about the energetics of the dissolution process and is a fundamental concept in understanding thermochemical reactions.

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Most popular questions from this chapter

Which of the following expression for an irreversible process: (a) \(d S>\frac{d q}{T}\) (b) \(d S=\frac{d q}{T}\) (c) \(d S<\frac{d q}{T}\) (d) \(d S=\frac{d U}{T}\)

\begin{aligned} &\text { Suppose that a reaction has } \Delta H=-40 \mathrm{~kJ} \text { and } \Delta S=-50 \mathrm{~J} / \mathrm{K} \text { . At what temperature range will it } \\ &\text { change from spontaneous to non-spontaneous? } \\ &\begin{array}{llll} \text { (a) } 0.8 \mathrm{~K} \text { to } 1 \mathrm{~K} & \text { (b) } 799 \mathrm{~K} \text { to } 800 \mathrm{~K} & \text { (c) } 800 \mathrm{~K} \text { to } 801 \mathrm{~K} & \text { (d) } 799 \mathrm{~K} \text { to } 801 \mathrm{~K} \end{array} \end{aligned}

With what minimum pressure (in \(k P a\) ), a given volume of an ideal gas \(\left(C_{p, m}=7 / 2 R\right.\) ), originally at \(400 \mathrm{~K}\) and \(100 \mathrm{kPa}\) pressure can be compressed irreversibly adiabatically in order to raise its temperature to \(600 \mathrm{~K}\) : (a) \(362.5 \mathrm{kPa}\) (b) \(275 \mathrm{kPa}\) (c) \(437.5 \mathrm{kPa}\) (d) \(550 \mathrm{kPa}\)

Calculate standard entropy change in the reaction $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ Given : \(S_{m}^{\circ}\left(\mathrm{Fe}_{2} \mathrm{O}_{3}, \mathrm{~S}\right)=87.4, S_{m}^{\circ}(\mathrm{Fe}, S)=27.3\) \(S_{m}^{\circ}\left(\mathrm{H}_{2}, g\right)=130.7, S_{m}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}, l\right)=69.9 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Consider the following reaction: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H_{1}=44 \mathrm{~kJ} \\ 2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) ; \Delta H_{2}=-1453 \mathrm{~kJ} \end{aligned} $$ What is the value of \(\Delta H\) for second reaction if water vapour instead of liquid water is formed as product? (a) \(-1409 \mathrm{~kJ}\) (b) \(-1629 \mathrm{~kJ}\) (c) \(-1277 \mathrm{~kJ}\) (d) None of these

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