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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 \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\), calculate 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 HCl by NaOH is \(56,116 \: J/mol\).

Step by step solution

01

1. Calculate the temperature difference

First, we need to determine the change in temperature during the reaction. To do this, subtract the initial temperature from the final temperature: ΔT = T_final - T_initial = 31.3°C - 24.6°C = 6.7°C
02

2. Calculate the mass of the mixture

To find the mass of the mixture, multiply the volume by the density: mass = volume x density = (100.0 mL HCl + 100.0 mL NaOH) x 1.0 g/cm³ = 200 g
03

3. Calculate the heat released (Q)

Now we will calculate the heat released during the reaction using the formula Q = mcΔT, where m is mass and c is the specific heat capacity. Q = mass x specific heat capacity x ΔT = 200 g x 4.18 J/°C·g x 6.7°C = 5611.6 J
04

4. Calculate the number of moles of HCl and NaOH

Next, we need to determine the number of moles of reactants. Since HCl and NaOH have the same volume and concentration, their mole amounts are equal. moles = concentration x volume = 1.0 mol/L x 0.100 L = 0.100 mol
05

5. Calculate the enthalpy change

Finally, we will find the enthalpy change for the neutralization reaction by dividing the total heat released (Q) by the number of moles of reactants involved in the reaction. Enthalpy change (ΔH) = Q / moles = 5611.6 J / 0.100 mol = 56116 J/mol The enthalpy change for the neutralization of HCl by NaOH is \(56,116 \: J/mol\).

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Key Concepts

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

Enthalpy Change
Enthalpy change, represented as \( \Delta H \), is a measure of the total heat content in a chemical reaction at constant pressure. In calorimetry, it's key to understanding how reactions release or absorb energy. During a reaction, the enthalpy change can illustrate whether the process is exothermic (releasing heat) or endothermic (absorbing heat). In our exercise, the neutralization of HCl by NaOH is exothermic, as the resulting temperature of the mixture rose from \(24.6^{\circ}\text{C}\) to \(31.3^{\circ}\text{C}\). This rise indicates that heat has been released into the surroundings.
Neutralization Reaction
A neutralization reaction occurs when an acid reacts with a base to form water and a salt. The general form is:\[ \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} \]In this exercise, hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH), creating sodium chloride (NaCl) and water (H\(_2\)O). The neutralization reaction is important in calorimetry because it typically releases energy as heat. The energy change in this reaction is detected by the temperature change in the solution, which helps us calculate the enthalpy change. In practical applications, understanding these reactions can aid in processes such as chemical manufacturing and waste treatment.
Specific Heat Capacity
Specific heat capacity is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (\(1^{\circ}\text{C}\)). It is represented by \(c\) and for water and solutions with properties similar to water, it is approximately \(4.18 \, \text{J/g}^{\circ}\text{C}\).In our scenario, this specific heat capacity helps us calculate the specific energy release when the mixtures are combined. Without this universal heat factor, our energy calculations would lack precision. Using specific heat capacity allows us to quantify the subtle differences in temperature changes during chemical reactions, which is crucial for determining the energy changes accurately.
Coffee-Cup Calorimeter
A coffee-cup calorimeter is a simple, yet effective tool for measuring the heat effects of reactions carried out at constant pressure, often in aqueous solutions. It typically consists of a polystyrene cup, a lid, a thermometer, and sometimes a stirring mechanism. This setup provides insulation, limiting the heat exchange between the reaction inside and the environment. In our exercise, using a coffee-cup calorimeter allows us to monitor the temperature change directly as the reaction proceeds.
  • The initial and final temperatures are easily measurable.
  • Minimal heat loss ensures accurate data for calculating enthalpy changes.
  • It simplifies practical experiments for educational purposes.
Such equipment is indispensable in chemistry labs for its efficiency in giving quick and relatively accurate measurements of heat changes in neutralization and other simple reactions.

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Most popular questions from this chapter

The enthalpy of neutralization for the reaction of a strong acid with a strong base is \(-56 \mathrm{~kJ} / \mathrm{mol}\) water produced. How much energy will be released when \(200.0 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{HCl}\) is mixed with \(150.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) ?

Water gas is produced from the reaction of steam with coal: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}(g) $$ Assuming that coal is pure graphite, calculate \(\Delta H^{\circ}\) for this reaction.

At \(298 \mathrm{~K}\), the standard enthalpies of formation for \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) and \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) are \(227 \mathrm{~kJ} / \mathrm{mol}\) and \(49 \mathrm{~kJ} / \mathrm{mol}\), respectively. a. Calculate \(\Delta H^{\circ}\) for $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) $$ b. Both acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) can be used as fuels. Which compound would liberate more energy per gram when combusted in air?

Calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ \begin{aligned} 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) & \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1010 . \mathrm{kJ} \\ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) & \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-317 \mathrm{~kJ} \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-143 \mathrm{~kJ} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \Delta H &=-286 \mathrm{~kJ} \end{aligned} $$

Consider the following reaction: $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{~kJ} $$ a. How much heat is evolved for the production of \(1.00 \mathrm{~mol}\) \(\mathrm{H}_{2} \mathrm{O}(l) ?\) b. How much heat is evolved when \(4.03 \mathrm{~g}\) hydrogen is reacted with excess oxygen? c. How much heat is evolved when \(186 \mathrm{~g}\) oxygen is reacted with excess hydrogen? d. The total volume of hydrogen gas needed to fill the Hindenburg was \(2.0 \times 10^{8} \mathrm{~L}\) at \(1.0 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\). How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?

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