When a \(6.50-\mathrm{g}\) sample of solid sodium hydroxide dissolves in \(100.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.18 ), the temperature rises from \(21.6^{\circ} \mathrm{C}\) to \(37.8^{\circ} \mathrm{C}\). Calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH})\) for the solution process $$ \mathrm{NaOH}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water.

Short Answer

Expert verified
The molar enthalpy change for the dissolution process of sodium hydroxide is \(\Delta H = 44.31 \: \mathrm{kJ/mol}\).

Step by step solution

01

Calculate the heat transferred during the dissolution process

To calculate the heat transferred (q) during the dissolution process, we will use the formula: q = mc∆T where m is the mass of the solution (m = mass of NaOH + mass of water), c is the specific heat capacity of the solution, ∆T is the change in temperature (T_final - T_initial) Given data: Mass of sodium hydroxide (NaOH) = 6.50 g Mass of water = 100.0 g Initial temperature (T_initial) = 21.6 °C Final temperature (T_final) = 37.8 °C Specific heat capacity (c) = 4.18 J/(g°C) Let's calculate the heat transferred (q) m = Mass of NaOH + Mass of water = 6.50 g + 100.0 g = 106.50 g ∆T = T_final - T_initial = 37.8 °C - 21.6 °C = 16.2 °C q = (106.50 g)(4.18 J/(g°C))(16.2 °C) = 7201.26 J
02

Calculate the moles of sodium hydroxide

To calculate the moles of sodium hydroxide (NaOH), we will use the formula: moles of NaOH = mass of NaOH/Molar mass of NaOH Given data: Mass of sodium hydroxide (NaOH) = 6.50 g Molar mass of sodium hydroxide (NaOH) = 22.99 g/mol (Na) + 15.99 g/mol (O) + 1.008 g/mol (H) = 40.00 g/mol Let's calculate the moles of sodium hydroxide (NaOH) moles of NaOH = 6.50 g / 40.00 g/mol = 0.1625 mol
03

Calculate the enthalpy change per mole of sodium hydroxide

To determine the molar enthalpy change for the dissolution process, we will use the formula: ΔH = q / moles of NaOH Given data: Heat transferred (q) = 7201.26 J Moles of sodium hydroxide (NaOH) = 0.1625 mol Let's calculate the molar enthalpy change (ΔH) ΔH = 7201.26 J / 0.1625 mol = 44,312.24 J/mol Since the question asks for the answer in kJ/mol, ΔH = 44.31 kJ/mol The molar enthalpy change for the dissolution process of sodium hydroxide is 44.31 kJ/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, commonly denoted as \( \Delta H \), represents the heat absorbed or released during a chemical reaction at constant pressure. It's an essential concept in thermodynamics and chemistry, as it helps predict whether a process will release or consume energy. An enthalpy change can be either endothermic (absorbing heat from the surroundings, \( \Delta H > 0 \) or exothermic (releasing heat to the surroundings, \( \Delta H < 0 \).

In our exercise, we observe the dissolution process of sodium hydroxide (NaOH) in water. This process absorbs heat, indicating an endothermic reaction, and the positive \( \Delta H \) confirms this. To calculate it step wise, we first determine the amount of heat transferred using calorimetry, then find the number of moles of NaOH dissolved, and finally, divide the heat value by the number of moles to get the enthalpy change per mole.
Dissolution Process
The dissolution process involves the breaking and forming of bonds as a solute dissolves in a solvent to form a solution. When solid sodium hydroxide dissolves in water, it dissociates into \( \mathrm{Na}^+ \) and \( \mathrm{OH}^- \) ions, a process that can absorb or release energy. The increase in the water's temperature upon the addition of NaOH suggests that the dissolution is an endothermic process - meaning it absorbs heat from the surroundings.

It is critical to consider that the solubility and the enthalpy change of a substance are dependent on its nature and the solvent's properties. In our case, the specific heat capacity is considered to be the same as that of water, which is a common assumption in calorimetry when dealing with aqueous solutions.
Calorimetry
Calorimetry is the measurement of heat from chemical reactions, phase transitions, or physical changes. A calorimeter is an instrument used for this purpose. In the context of our exercise, we relied on a coffee-cup calorimeter, a simple and common tool in educational labs for measuring heat changes.

The process requires us to calculate the heat absorbed or released (\( q \) using the equation \( q = mc\Delta T \), where \( m \) is the total mass of the solution, \( c \) is the specific heat capacity (assuming the value for pure water), and \( \Delta T \) is the change in temperature. Clear step-by-step calculations demonstrate how calorimetry aids in determining the enthalpy change during the dissolution of sodium hydroxide in water.

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Most popular questions from this chapter

Three common hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: (a) For each of these substances, calculate the molar enthalpy of combustion to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Calculate the fuel value in \(\mathrm{kJ} / \mathrm{g}\) for each of these compounds. \((\mathrm{c})\) For each hydrocarbon, determine the percentage of hydrogen by mass. (d) By comparing your answers for parts (b) and (c), propose a relationship between hydrogen content and fuel value in hydrocarbons.

Using values from Appendix \(\mathrm{C},\) calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(\mathrm{CaO}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{CaCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (c) \(2 \mathrm{CuO}(s)+\mathrm{NO}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{NO}_{2}(g)\) (d) \(4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Imagine that you are climbing a mountain. (a) Is the distance you travel to the top a state function? Why or why not? (b) Is the change in elevation between your base camp and the peak a state function? Why or why not? [Section 5.2]

Ozone, \(\mathrm{O}_{3}(g)\), is a form of elemental oxygen that is important in the absorption of ultraviolet radiation in the stratosphere. It decomposes to \(\mathrm{O}_{2}(g)\) at room temperature and pressure according to the following reaction: $$ 2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) \quad \Delta H=-284.6 \mathrm{~kJ} $$ (a) What is the enthalpy change for this reaction per mole of \(\mathrm{O}_{3}(g) ?(\mathbf{b})\) Which has the higher enthalpy under these condi- $$ \text { tions, } 2 \mathrm{O}_{3}(g) \text { or } 3 \mathrm{O}_{2}(g) ? $$

Using values from Appendix \(\mathrm{C},\) calculate the standard enthalpy change for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)\)

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