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Chapter 7: Problem 27

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) torches are used in welding. How much heat (in kJ) evolves when 5.0 L of \(C_{2} \mathrm{H}_{2}\) \(\left(d=1.0967 \mathrm{kg} / \mathrm{m}^{3}\right)\) is mixed with a stoichiometric amount of oxygen gas? The combustion reaction is $$\begin{array}{r} \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-1299.5 \mathrm{kJ} \end{array}$$

Short Answer

Expert verified
The heat evolved when 5.0 L of acetylene is mixed in a stoichiometric amount of oxygen gas is -273.8 kJ

Step by step solution

01

Determine the molecular weight of acetylene

The molecular weight of \(\mathrm{C}_{2} \mathrm{H}_{2}\) is computed by adding the atomic masses of Carbon and Hydrogen. Given that the atomic mass of Carbon (C) is 12 g/mol and that of Hydrogen (H) is 1 g/mol, the molecular weight of \(\mathrm{C}_{2} \mathrm{H}_{2}\) is \(= 2 \times 12 (Carbon) + 2 \times 1 (Hydrogen) = 26 \, \mathrm{g/mol}\)
02

Determine mass of acetylene

We are given the volume of \(\mathrm{C}_{2} \mathrm{H}_{2}\) to be 5.0 L and its density \(d = 1.0967 \mathrm{kg/m^{3}} = 1.0967 \mathrm{g/cm^{3}}\). Density is \(mass/volume\). Therefore, we calculate mass by multiplying volume by density. So, the mass of acetylene is \(mass = volume \times density = 5.0 L \times 1.0967 \mathrm{g/cm^{3}} = 5.4835 g\)
03

Find the moles of acetylene

Moles can be calculated using the formula \(moles = mass/molecular \, weight\). So, the moles of acetylene, \(moles_{C2H2} = mass_{C2H2} / molecular \, weight_{C2H2} = 5.4835 g / 26 g/mol = 0.2109 mol\)
04

Determine heat evolved

The combustion reaction shows that 1 mol of \(\mathrm{C}_{2} \mathrm{H}_{2}\) releases \(-1299.5 kJ\). Therefore, \(0.2109 mol\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) will release \(∆H = 0.2109 mol \times -1299.5 kJ/mol = -273.8 kJ\)

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Most popular questions from this chapter

Substitute natural gas (SNG) is a gaseous mixture containing \(\mathrm{CH}_{4}(\mathrm{g})\) that can be used as a fuel. One reaction for the production of SNG is $$\begin{aligned} 4 \mathrm{CO}(\mathrm{g})+8 \mathrm{H}_{2}(\mathrm{g}) & \longrightarrow \\ 3 \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & \Delta H^{\circ}=? \end{aligned}$$ Use appropriate data from the following list to determine \(\Delta H^{\circ}\) for this SNG reaction. $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ \quad\quad\quad\quad\quad\quad\quad\quad\qquad \Delta H^{\circ}=-110.5 \mathrm{k} \mathrm{J} \end{array}$$$$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-283.0 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{array}{l} \text { C(graphite) }+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g}) \\ \qquad \Delta H^{\circ}=-74.81 \mathrm{kJ} \end{array}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ & \Delta H^{\circ}=-890.3 \mathrm{kJ} \end{aligned}$$

We can determine the purity of solid materials by using calorimetry. A gold ring (for pure gold, specific heat \(=0.1291 \mathrm{Jg}^{-1} \mathrm{K}^{-1}\) ) with mass of \(10.5 \mathrm{g}\) is heated to \(78.3^{\circ} \mathrm{C}\) and immersed in \(50.0 \mathrm{g}\) of \(23.7^{\circ} \mathrm{C}\) water in a constant-pressure calorimeter. The final temperature of the water is \(31.0^{\circ} \mathrm{C}\). Is this a pure sample of gold?

An alternative approach to bomb calorimetry is to establish the heat capacity of the calorimeter, exclusive of the water it contains. The heat absorbed by the water and by the rest of the calorimeter must be calculated separately and then added together. A bomb calorimeter assembly containing \(983.5 \mathrm{g}\) water is calibrated by the combustion of \(1.354 \mathrm{g}\) anthracene. The temperature of the calorimeter rises from 24.87 to \(35.63^{\circ} \mathrm{C} .\) When \(1.053 \mathrm{g}\) citric acid is burned in the same assembly, but with 968.6 g water, the temperature increases from 25.01 to \(27.19^{\circ} \mathrm{C}\). The heat of combustion of anthracene, \(\mathrm{C}_{14} \mathrm{H}_{10}(\mathrm{s}),\) is \(-7067 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{C}_{14} \mathrm{H}_{10} \cdot\) What is the heat of combustion of citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7},\) expressed in \(\mathrm{kJ} / \mathrm{mol} ?\)

Explain the important distinctions between each pair of terms: (a) system and surroundings; (b) heat and work; (c) specific heat and heat capacity; (d) endothermic and exothermic; (e) constant-volume process and constant-pressure process.

In the Are You Wondering \(7-1\) box, the temperature variation of enthalpy is discussed, and the equation \(q_{P}=\) heat capacity \(\times\) temperature change \(=C_{P} \times \Delta T\) was introduced to show how enthalpy changes with temperature for a constant-pressure process. Strictly speaking, the heat capacity of a substance at constant pressure is the slope of the line representing the variation of enthalpy (H) with temperature, that is $$C_{P}=\frac{d H}{d T} \quad(\text { at constant pressure })$$ where \(C_{P}\) is the heat capacity of the substance in question. Heat capacity is an extensive quantity and heat capacities are usually quoted as molar heat capacities \(C_{P, \mathrm{m}},\) the heat capacity of one mole of substance; an intensive property. The heat capacity at constant pressure is used to estimate the change in enthalpy due to a change in temperature. For infinitesimal changes in temperature, $$d H=C_{p} d T \quad(\text { at constant pressure })$$ To evaluate the change in enthalpy for a particular temperature change, from \(T_{1}\) to \(T_{2}\), we write $$\int_{H\left(T_{1}\right)}^{H\left(T_{2}\right)} d H=H\left(T_{2}\right)-H\left(T_{1}\right)=\int_{T_{1}}^{T_{2}} C_{P} d T$$ If we assume that \(C_{P}\) is independent of temperature, then we recover equation (7.5) $$\Delta H=C_{P} \times \Delta T$$ On the other hand, we often find that the heat capacity is a function of temperature; a convenient empirical expression is $$C_{P, \mathrm{m}}=a+b T+\frac{c}{T^{2}}$$ What is the change in molar enthalpy of \(\mathrm{N}_{2}\) when it is heated from \(25.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} ?\) The molar heat capacity of nitrogen is given by$$C_{P, \mathrm{m}}=28.58+3.77 \times 10^{-3} T-\frac{0.5 \times 10^{5}}{T^{2}} \mathrm{JK}^{-1} \mathrm{mol}^{-1}$$
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