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When \(0.100 \mathrm{~g} \mathrm{CaO}(\mathrm{s})\) is added to \(125 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) at \(23.6^{\circ} \mathrm{C}\) in a coffee cup calorimeter, this reaction occurs. \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq}) \quad \Delta_{t} H^{\circ}=-81.9 \mathrm{~kJ} / \mathrm{mol}\) Calculate the final temperature of the solution.

Short Answer

Expert verified
The final temperature of the solution is approximately 23.32°C.

Step by step solution

01

Calculate Moles of CaO

First, we need to calculate the number of moles of calcium oxide (CaO) using its mass and molar mass. The molar mass of CaO is approximately 56.08 g/mol. Use the formula:\[\text{moles of CaO} = \frac{\text{mass of CaO}}{\text{molar mass of CaO}} = \frac{0.100\, \text{g}}{56.08\, \text{g/mol}}\]This calculation yields approximately 0.00178 moles of CaO.
02

Calculate Heat Released

Using the number of moles of CaO and the reaction enthalpy, calculate the heat released by the reaction. The reaction enthalpy \(\Delta H\) is given as \(-81.9\, \text{kJ/mol}\). Use the formula:\[q = \text{moles of CaO} \times \Delta H = 0.00178\, \text{mol} \times (-81.9\, \text{kJ/mol})\]This results in approximately \(-0.1458\, \text{kJ}\) or \(-145.8\, \text{J}\) of heat released.
03

Calculate Heat Absorbed by Water

The heat released by the reaction is absorbed by the water. Use the formula for heat:\[q = mc\Delta T\]where \(m\) is the mass of water (125 g), \(c\) is the specific heat capacity of water (4.18 J/g°C), and \(\Delta T\) is the change in temperature. Set the heat absorbed by water equal to the negative of the heat released:\[mc\Delta T = -145.8\, \text{J}\]\[125\, \text{g} \times 4.18\, \text{J/g°C} \times \Delta T = -145.8\, \text{J}\]
04

Solve for Temperature Change

Solve \(125 \times 4.18 \times \Delta T = -145.8\, \text{J}\) for \(\Delta T\):\[\Delta T = \frac{-145.8}{125 \times 4.18} \approx -0.28\, ^\circ \text{C}\]This indicates the solution's temperature change is a decrease of about 0.28°C.
05

Calculate Final Temperature

Use the initial temperature and the temperature change to find the final temperature:\[T_{\text{final}} = T_{\text{initial}} + \Delta T = 23.6\, ^\circ \text{C} - 0.28\, ^\circ \text{C}\]So, the final temperature of the solution is approximately 23.32°C.

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Key Concepts

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

Enthalpy Change
In chemical reactions, the change in enthalpy (\( \Delta H \)) represents the heat absorbed or released. For this specific reaction:
  • Enthalpy change (\( \Delta_{t} H^{\circ} \)) is \(-81.9 \, \text{kJ/mol}\)
  • Negative value means the reaction releases heat (exothermic)
Understanding \( \Delta H \) helps predict if a reaction will warm or cool its surroundings. Here, the reaction between \( \text{CaO} \) and \( \text{H}_2\text{O} \) releases heat, indicating it's exothermic. In calorimetry, this energy transfer guides us in calculating temperature changes.
Specific Heat Capacity
Specific heat capacity (\( c \)) is the measure of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. For water:
  • Specific heat capacity \( c = 4.18 \, \text{J/g°C} \)
  • High capacity makes water effective for absorbing and releasing heat
Knowing the specific heat capacity helps in determining how much heat a substance, like water, can absorb before changing temperature. This property is essential in calorimetry experiments, as it influences how the entire system responds to heat flows.
Temperature Change
Temperature change (\( \Delta T \)) reflects how much the temperature of a system shifts due to heat exchange. In our exercise, the water absorbs heat released by the reaction, leading to:
  • Temperature decrease of \(-0.28 \, ^\circ \text{C}\)
  • Calculated from \( mc\Delta T = q \)
This change is found using mass, specific heat capacity, and the heat exchanged. The calculation ensures energy conservation within the calorimetry process, maintaining balance between heat absorbed and released.
Chemical Reactions in Solutions
Chemical reactions in solutions often involve substances dissolving or changing composition. The reaction of \( \text{CaO} \) and \( \text{H}_2\text{O} \) forms a new product (\( \text{Ca(OH)}_2 \)). Key aspects include:
  • Transformation into an aqueous solution
  • Energy exchange, impacting temperature
These reactions are central in calorimetry as they illustrate energy changes efficiently. Solvent properties, like water’s high capacity, ensure heat movements are visible via temperature shifts. They are core to understanding chemical dynamics in liquid mediums.

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Most popular questions from this chapter

"Gasohol," a mixture of gasoline and ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) is used as automobile fuel. The alcohol releases energy in a combustion reaction with \(\mathrm{O}_{2}\). $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\ell) $$ If \(0.115 \mathrm{~g}\) ethanol evolves \(3.62 \mathrm{~kJ}\) when burned at constant pressure, calculate the combustion enthalpy for ethanol.

A \(25.0-\mathrm{mL}\) sample of benzene at \(19.9{ }^{\circ} \mathrm{C}\) was cooled to its melting point, \(5.5^{\circ} \mathrm{C}\), and then frozen. Calculate how much heat transfer to the surroundings occurred in this process. (The density of benzene is \(0.880 \mathrm{~g} / \mathrm{mL} ;\) its specific heat capacity is \(1.74 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}\), and its enthalpy of fusion is \(127 \mathrm{~J} / \mathrm{g} .)\)

Melting lead requires 5.50 cal/g. Calculate how many joules are required to melt \(1.00 \mathrm{lb}\) (454 g) lead.

These reaction enthalpies can be measured: \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) \(\Delta_{\mathrm{r}} H^{\circ}=-1411.1 \mathrm{~kJ} / \mathrm{mol}\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\ell)\) \(\Delta_{t} H^{\circ}=-1367.5 \mathrm{~kJ} / \mathrm{mol}\) Use these values and Hess's law to determine the reaction enthalpy for $$ \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell) $$

You wish to know the standard formation enthalpy of liquid \(\mathrm{PCl}_{3}\) $$ \mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{3}(\ell) $$ These reaction enthalpies have been determined experimentally: $$ \begin{array}{ll} \mathrm{P}_{4}(\mathrm{~s})+10 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{5}(\mathrm{~s}) & \Delta_{\mathrm{r}} H^{\circ}=-1774.0 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{PCl}_{3}(\ell)+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{~s}) & \Delta_{\mathrm{r}} H^{\circ}=-123.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Calculate the formation enthalpy for \(\mathrm{PCl}_{3}(\ell)\).

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