In a coffee-cup calorimeter, \(100.0 \mathrm{~mL}\) of \(1.0 \mathrm{M} \mathrm{NaOH}\) and \(100.0 \mathrm{~mL}\) of \(1.0 \mathrm{M} \mathrm{HCl}\) are mixed. Both solutions were originally at \(24.6^{\circ} \mathrm{C}\). After the reaction, the final temperature is \(31.3^{\circ} \mathrm{C}\). Assuming that all the solutions have a density of \(1.0 \mathrm{~g} / \mathrm{cm}^{3}\) and a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\), calcu- late the enthalpy change for the neutralization of \(\mathrm{HCl}\) by \(\mathrm{NaOH}\). Assume that no heat is lost to the surroundings or to the calorimeter.

Short Answer

Expert verified
The enthalpy change for the neutralization of \(\mathrm{HCl}\) by \(\mathrm{NaOH}\) is \(-56119.6 \mathrm{~J/mol}\).

Step by step solution

01

Calculate the moles of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\)

The solution of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\) both have 1.0 M concentration, and they have equal volumes (100.0 mL). Thus, they have the same number of moles. The moles can be found using this formula: moles = Volume (in L) × Molarity. Moles of \(\mathrm{HCl}\) = Moles of \(\mathrm{NaOH}\) = \(0.100 \mathrm{~L} \times 1.0 \mathrm{~M} = 0.100 \mathrm{~mol}\).
02

Calculate the heat gained by the solutions

The heat gained by the solutions is equal to the mass × specific heat capacity × change in temperature. Since the mass of both solutions is the same, we can find the combined mass by multiplying the volume and the density. Combined mass = Volume of one solution × Density × 2 = \(100.0 \mathrm{~mL} \times 1.0 \mathrm{~g/cm^3} \times 2 = 200.0 \mathrm{~g}\) Next, calculate the change in temperature: \(\Delta T = T_{final} - T_{initial} = 31.3^{\circ} \mathrm{C} - 24.6^{\circ} \mathrm{C} = 6.7^{\circ} \mathrm{C}\) Now, we can calculate the heat gained by the solutions: \(Q_{gained} = m \times C \times \Delta T = 200.0 \mathrm{~g} \times 4.18 \mathrm{~J/^{\circ}C \cdot g} \times 6.7^{\circ} \mathrm{C} = 5611.96 \mathrm{~J}\)
03

Calculate the heat released by the reaction

By conservation of energy, the heat released by the reaction is equal to the heat gained by the solutions (as we assumed that no heat is lost to the surroundings or the calorimeter). \(Q_{released} = Q_{gained} = 5611.96 \mathrm{~J}\)
04

Calculate the enthalpy change for the neutralization reaction

The enthalpy change for the neutralization reaction is the heat released divided by the moles of \(\mathrm{HCl}\) (or \(\mathrm{NaOH}\) as they have equal moles). \(\Delta H = \frac{Q_{released}}{\text{moles of } \mathrm{HCl}} = -\frac{5611.96 \mathrm{~J}}{0.100 \mathrm{~mol}} = -56119.6 \mathrm{~J/mol}\) Here, \(\Delta H\) is negative, which indicates the reaction is exothermic. The enthalpy change for the neutralization of \(\mathrm{HCl}\) by \(\mathrm{NaOH}\) is -56119.6 \(\mathrm{J/mol}\).

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Key Concepts

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

Calorimetry
Calorimetry is a technique used to measure the amount of heat involved in a chemical or physical process. This technique is crucial for understanding endothermic and exothermic reactions, including neutralization reactions, which are fundamental concepts in thermodynamics and chemistry.

In our exercise, a coffee-cup calorimeter was used, which is a simple calorimeter composed of two nested Styrofoam cups, a cover, and often includes a thermometer. The purpose of the calorimeter is to provide insulation, ensuring that heat exchange occurs only between the substances inside the cup and not with the external environment. To determine the heat exchange, it is assumed that no heat is lost to the surroundings.

To calculate the enthalpy change, one must consider the specific heat capacity of the solutions, the mass of the solutions, and the temperature change. The specific heat capacity indicates how much heat is required to raise the temperature of one gram of a substance by one degree Celsius. The mass is the combined weight of the two solutions. Finally, the temperature change (\f\(\triangle T\f\)) is found by subtracting the initial temperature from the final temperature. This information is used to calculate the heat emitted or absorbed during the reaction (\f\(Q\f\)), which then allows for calculation of the enthalpy change.
Neutralization Reactions
Neutralization reactions are chemical reactions in which an acid and a base react to form water and a salt. This type of reaction is extremely important in various fields such as chemistry, biochemistry, and environmental science.

In the context of our exercise, the neutralization involves hydrochloric acid (\f\(HCl\f\)) and sodium hydroxide (\f\(NaOH\f\)), both strong acid and base, respectively. When they react, the hydrogen ions (\f\(H^+\f\)) from the acid and the hydroxide ions (\f\(OH^-\f\)) from the base combine to form water (\f\(H_2O\f\)), releasing an amount of heat in the process. This heat is what we measure using calorimetry.

This release or absorption of heat is referred to as the enthalpy change (\f\(\triangle H\f\)), which is often expressed in joules per mole (\f\(J/mol\f\)). The sign of the enthalpy change denotes whether the reaction is exothermic (negative sign, releasing heat) or endothermic (positive sign, absorbing heat). The exercise presents an exothermic neutralization reaction, as evidenced by the negative enthalpy value obtained from the calculations.
Heat Capacity
Heat capacity is an intrinsic property of a substance that describes the amount of heat required to change its temperature by a given amount. The specific heat capacity, often just called specific heat, is the amount of heat per unit mass required to raise the temperature by one degree Celsius (\f\(J/^{\text{o}}C \bullet g\f\)).

From our calculations in the exercise, the specific heat capacity was used to determine the heat gained by the solutions during the neutralization reaction. Knowing the specific heat capacity of the solutions allows us to calculate the amount of heat energy absorbed or released.

  • The specific heat capacity of the solutions (\f\(4.18 J/^{\text{o}}C \bullet g\f\)) is a key variable in our calorimetry equation.
  • It's noteworthy that water, the primary component of the solutions in the exercise, has a high specific heat capacity, which allows it to absorb or release significant amounts of heat with little change in temperature.

Understanding specific heat capacity is essential because it explains the temperature stability of substances and the energy exchange during thermal processes, such as those observed in neutralization reactions.

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Most popular questions from this chapter

The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ a. Use the values of \(\Delta H_{\mathrm{f}}^{\circ}\) in Appendix 4 to calculate the value of \(\Delta H^{\circ}\) for each of the preceding reactions. b. Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.) Is the overall reaction exothermic or endothermic?

A swimming pool, \(10.0 \mathrm{~m}\) by \(4.0 \mathrm{~m}\), is filled with water to a depth of \(3.0 \mathrm{~m}\) at a temperature of \(20.2^{\circ} \mathrm{C}\). How much energy is required to raise the temperature of the water to \(24.6^{\circ} \mathrm{C} ?\)

Consider the following changes: a. \(\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)\) b. \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)\) d. \(2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) e. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)\) At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

If the internal energy of a thermodynamic system is increased by \(300 . \mathrm{J}\) while \(75 \mathrm{~J}\) of expansion work is done, how much heat was transferred and in which direction, to or from the system?

A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing \(150.0 \mathrm{~g}\) water at \(23.5^{\circ} \mathrm{C}\). After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C}\). If the specific heat capacity of nickel is \(0.444 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\), what mass of nickel was originally heated? Assume no heat loss to the surroundings.

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