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Chapter 5: Problem 56

(a) When a 4.25 -g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter (Figure 5.18), the temperature drops from 22.0 to \(16.9^{\circ} \mathrm{C}\) . Calculate \(\Delta H\left(\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) for the solution process: $$\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)$$ Assume that the specific heat of the solution is the same as that of pure water. (b) Is this process endothermic or exothermic?

Short Answer

Expert verified
The enthalpy change for the dissolution of NH₄NO₃ is \(-24.065\,\text{kJ/mol}\), and the process is exothermic.

Step by step solution

01

List the given information

- Mass of NH₄NO₃ (m₁) = 4.25g - Mass of H₂O (m₂) = 60.0g - Initial temperature (T₁) = 22.0°C - Final temperature (T₂) = 16.9°C - Specific heat capacity of water (c) = 4.18 J/(g°C) Note: The specific heat capacity of the solution is assumed to be the same as that of pure water.
02

Calculate the temperature change (∆T)

We will first calculate the temperature change by subtracting the final temperature from the initial temperature. ∆T = T₂ - T₁ = 16.9°C - 22.0°C = -5.1°C
03

Calculate the energy change (q)

Now, we will determine the energy change (q) using the equation q = mc∆T. Since the specific heat capacity of the solution is the same as that of pure water, we can use the mass of water (m₂) in our calculations. q = m₂ × c × ∆T = 60.0g × 4.18 J/(g°C) × (-5.1°C) q = -1277.64 J
04

Convert the mass of NH₄NO₃ to moles

To find the moles of NH₄NO₃, we will divide the mass (m₁) by the molar mass. Molar mass of NH₄NO₃ = (1 * 14.01 + 4 * 1.01) + (1 * 14.01 + 3 * 16.00) = 18.05 + 62.01 = 80.06 g/mol moles of NH₄NO₃ = m₁ / (molar mass) = 4.25g / 80.06 g/mol moles of NH₄NO₃ = 0.0531 mol
05

Calculate ∆H

Now, we will find ∆H by dividing the energy change (q) by the moles of NH₄NO₃ and converting the result to kJ/mol. ∆H = (q) / (moles of NH₄NO₃) = -1277.64 J / 0.0531 mol ∆H = -24,065 J/mol Since we need to convert the result to kJ/mol, divide by 1000: ∆H = -24.065 kJ/mol
06

Determine if the process is endothermic or exothermic

The value of ∆H is negative, which means the process is exothermic. This indicates that heat is released during the dissolution of NH₄NO₃. In conclusion, the enthalpy change for the dissolution of NH₄NO₃ is -24.065 kJ/mol, and the process is exothermic.

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Most popular questions from this chapter

Burning methane in oxygen can produce three different carbon-containing products: soot (very fine particles of graphite), CO(g), and \(\mathrm{CO}_{2}(g) .\) (a) Write three balanced equations for the reaction of methane gas with oxygen to produce these three products. In each case assume that \(\mathrm{H}_{2} \mathrm{O}(l)\) is the only other product. (b) Determine the standard enthalpies for the reactions in part (a).(c) Why, when the oxygen supply is adequate, is \(\mathrm{CO}_{2}(g)\) the predominant carbon-containing product of the combustion of methane?

A 2.200 -g sample of quinone \(\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)\) is burned in a bomb calorimeter whose total heat capacity is 7.854 \(\mathrm{kJ} / \mathrm{c}\) . The temperature of the calorimeter increases from 23.44 to \(30.57^{\circ} \mathrm{C}\) . What is the heat of combustion per gram of quinone? Per mole of quinone?

The standard enthalpies of formation of gaseous propyne \(\left(\mathrm{C}_{3} \mathrm{H}_{4}\right),\) propylene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right),\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) are \(+185.4,+20.4,\) and \(-103.8 \mathrm{kJ} / \mathrm{mol}\) , respectively.(a) Calculate the heat evolved per mole on combustion of each substance to yield \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) (b) Calculate the heat evolved on combustion of 1 \(\mathrm{kg}\) of each substance. (c) Which is the most efficient fuel in terms of heat evolved per unit mass?

The complete combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l),\) to form \(\mathrm{H}_{2} \mathrm{O}(g)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases 1235 \(\mathrm{kJ}\) of heat per mole of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)$$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$\begin{array}{ll}{\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-890.3 \mathrm{kJ}} \\ {\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)} & {\Delta H^{\circ}=-136.3 \mathrm{kJ}} \\ {2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-571.6 \mathrm{kJ}} \\ {2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-3120.8 \mathrm{kJ}}\end{array}$$
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