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The heat of solution of \(\mathrm{KI}(\mathrm{s})\) in water is \(+20.3 \mathrm{kJ} / \mathrm{mol}\) KI. If a quantity of KI is added to sufficient water at \(23.5^{\circ} \mathrm{C}\) in a Styrofoam cup to produce \(150.0 \mathrm{mL}\) of 2.50 M KI, what will be the final temperature? (Assume a density of \(1.30 \mathrm{g} / \mathrm{mL}\) and a specific heat of \(2.7 \mathrm{Jg}^{-1}\) \(\left.^{\circ} \mathrm{C}^{-1} \text {for } 2.50 \mathrm{M} \mathrm{KI} .\right)\)

Short Answer

Expert verified
The final temperature of the solution will be 37.8 °C.

Step by step solution

01

Calculate the moles of \(\mathrm{KI}\)

Firstly, we need to determine the amount of \(\mathrm{KI}\) in moles. To do so, multiply the volume of the solution by its molarity: \(n = V \cdot M\) where \(V = 150 \, mL = 0.150 \, L\) and \(M = 2.5 \, M\). Thus, \(n = 0.150 \, L \cdot 2.5 \, mol/L = 0.375 \, mol\).
02

Calculate the heat of solution Q

The heat of solution of \(\mathrm{KI}\) is given as \(+20.3 \, kJ/mol\). Therefore, the total heat produced when \(\mathrm{KI}\) dissolves can be calculated by multiplying the heat of solution per mole by the number of moles dissolved: \(Q = +20.3 \, kJ/mol \cdot 0.375 \, mol = +7.613 \, kJ\). The positive sign indicates this is an endothermic process, so the solution absorbs heat from its surroundings.
03

Convert Q to J

The value of Q obtained in the previous step should be converted from kJ to J because the specific heat is given in J/g°C. \(Q = +7.613 \, kJ = +7613 \, J\).
04

Calculate the mass of the solution

The mass (m) of the solution can be calculated by multiplying the volume of the solution by its density: \(m = V \cdot d = 150 \, mL \cdot 1.30 \, g/mL = 195 \, g\).
05

Determine \(\Delta T\) (change in temperature)

Use the formula \(q=mc\Delta T\) to find \( \Delta T\', which is \( \Delta T=q/(mc)\). The total heat Q=7613 J, the mass m of the solution is 195 g, and the specific heat c is 2.7 J/g°C. So, \( \Delta T = 7613 J / (195 g \cdot 2.7 J/g°C) = \Delta T = 14.3 °C ).
06

Calculate the final temperature

The final temperature is calculated by summing the initial temperature with the change in temperature: \(T_{final} = T_{initial} + \Delta T = 23.5 °C + 14.3 °C = 37.8 °C\).

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Key Concepts

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

Enthalpy Change
Enthalpy change is a measure of the total energy absorbed or released during a chemical reaction at constant pressure. In the context of the heat of solution, it signifies the energy change when a solute dissolves in a solvent. For the exercise we described, the quantity is given as a positive value of +20.3 kJ/mol for KI. This indicates that energy is absorbed from the surroundings.

When dealing with such processes, it is essential to consider whether the enthalpy change is positive or negative. A positive enthalpy change means the process is endothermic, requiring an input of energy. Conversely, a negative change signifies exothermic reactions, where energy is released.
  • The enthalpy change helps predict how temperature varies in chemical solutions.
  • Note that enthalpy is often measured in kilojoules per mole (kJ/mol).
Understanding enthalpy change is crucial in determining whether the surrounding environment will heat up or cool down during the solution process. Thus, it plays a fundamental role in forecasting the behavior of the reaction.
Endothermic Process
An endothermic process is one in which the system absorbs energy from its surroundings. This absorption usually manifests as a temperature decrease in the surroundings because the system is taking in heat. For our KI solution, the positive enthalpy change implies it's an endothermic process. As KI dissolves in water, energy is drawn from the environment, making the surroundings cooler initially.

It's important to recognize when a process is endothermic:
  • Endothermic reactions require energy to proceed, often sourced from the environment.
  • Typical signs of such processes include a temperature drop in the surrounding area.
In this exercise, as KI dissolves, the system absorbs 7613 J of energy. This justifies why the process feels cool to the touch initially. Eventually, however, since a protective Styrofoam cup isolate the system, the equilibrium restores, resulting in the observed final temperature increase to 37.8°C.
Specific Heat Capacity
Specific heat capacity is a fundamental concept in understanding how different substances react to heat energy. It is defined as the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. In the given problem, the specific heat capacity of the KI solution is 2.7 J/g°C. This relatively low specific heat implies that only a small amount of energy is needed to cause a relatively large temperature change.

To apply this in calculations, you must understand that:
  • Specific heat capacity is unique to each material.
  • A low specific heat means the substance heats up and cools down rapidly.
  • Specific heat is typically measured in joules per gram per degree Celsius (J/g°C).
In the exercise, the small specific heat capacity of 2.7 J/g°C allows for significant temperature changes with relatively little heat exchange. This is why the absorbed 7613 J of energy results in a noticeable final temperature increase from 23.5°C to 37.8°C. By understanding this value, you predict how much energy will affect temperature changes.

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Most popular questions from this chapter

A 0.75 g sample of \(\mathrm{KCl}\) is added to \(35.0 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) in a Styrofoam cup and stirred until it dissolves. The temperature of the solution drops from 24.8 to \(23.6^{\circ} \mathrm{C}\) (a) Is the process endothermic or exothermic? (b) What is the heat of solution of KCl expressed in kilojoules per mole of KCl?

Thermite mixtures are used for certain types of welding, and the thermite reaction is highly exothermic. $$\begin{array}{r} \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Fe}(\mathrm{s}) \\ \Delta H^{\circ}=-852 \mathrm{kJ} \end{array}$$ \(1.00 \mathrm{mol}\) of granular \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(2.00 \mathrm{mol}\) of granular Al are mixed at room temperature \(\left(25^{\circ} \mathrm{C}\right),\) and a reaction is initiated. The liberated heat is retained within the products, whose combined specific heat over a broad temperature range is about \(0.8 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} .\) (The melting point of iron is \(1530^{\circ} \mathrm{C} .\) ) Show that the quantity of heat liberated is more than sufficient to raise the temperature of the products to the melting point of iron.

A clay pot containing water at \(25^{\circ} \mathrm{C}\) is placed in the shade on a day in which the temperature is \(30^{\circ} \mathrm{C} .\) The outside of the clay pot is kept moist. Will the temperature of the water inside the clay pot (a) increase; (b) decrease; (c) remain the same?

Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction.Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction. (a) \(0.584 \mathrm{g}\) of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) yields \(29.4 \mathrm{kJ}\) (b) \(0.136 \mathrm{g}\) of camphor, \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}(\mathrm{s}),\) yields \(5.27 \mathrm{kJ}\) (c) \(2.35 \mathrm{mL}\) of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(\mathrm{l})(d=0.791\) \(\mathrm{g} / \mathrm{mL}),\) yields \(58.3 \mathrm{kJ}\)

Substitute natural gas (SNG) is a gaseous mixture containing \(\mathrm{CH}_{4}(\mathrm{g})\) that can be used as a fuel. One reaction for the production of SNG is $$\begin{aligned} 4 \mathrm{CO}(\mathrm{g})+8 \mathrm{H}_{2}(\mathrm{g}) & \longrightarrow \\ 3 \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & \Delta H^{\circ}=? \end{aligned}$$ Use appropriate data from the following list to determine \(\Delta H^{\circ}\) for this SNG reaction. $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ \quad\quad\quad\quad\quad\quad\quad\quad\qquad \Delta H^{\circ}=-110.5 \mathrm{k} \mathrm{J} \end{array}$$$$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-283.0 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{array}{l} \text { C(graphite) }+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g}) \\ \qquad \Delta H^{\circ}=-74.81 \mathrm{kJ} \end{array}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ & \Delta H^{\circ}=-890.3 \mathrm{kJ} \end{aligned}$$

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