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Chapter 7: Problem 43

Refer to Example \(7-3 .\) Based on the heat of combustion of sucrose established in the example, what should be the temperature change \((\Delta T)\) produced by the combustion of \(1.227 \mathrm{g} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) in a bomb calorimeter assembly with a heat capacity of \(3.87 \mathrm{kJ} /^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The temperature change \(\Delta T\) produced by the combustion of 1.227g sucrose in a bomb calorimeter assembly with a heat capacity of 3.87kJ/degC is approximately 4.26 degrees Celsius.

Step by step solution

01

Analyze the Information Given

We know the heat capacity \(C\) of the bomb calorimeter is \(3.87 \, \mathrm{kJ/ ^{\circ}C}\). We also know the mass \(m\) of the sucrose (\(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}\)) combusted is \(1.227 \, \mathrm{g}\). The molar mass of sucrose is approximately 342.30 g/mol. The heat of combustion is the heat released when one mole of a substance is completely burned in excess oxygen. From Example 7-3 we know that the heat of combustion \(q_p\) for sucrose is -5648 kJ/mol.
02

Calculate the Amount of Sucrose Combusted

We calculate the number of moles \(n\) of sucrose that has been combusted using the formula: \(n = \frac{m_i}{MM} = \frac{1.227\, \mathrm{g}}{342.3\, \mathrm{g/mol}}\), where \(m_i\) is the initial mass of the sucrose and \(MM\) is the molar mass.
03

Calculate the Heat of Combustion for the Given Mass

We calculate the heat \(q_p\) resulting from the combustion of the given amount of sucrose using the formula \(-q_p = n \cdot \Delta H = 1.227\, \mathrm{g} \cdot \frac{-5648\, \mathrm{kJ/mol}}{342.3\, \mathrm{g/mol}}\). This is the heat that is transferred to the calorimeter.
04

Calculate Change in Temperature

Finally, calculate the temperature change \(ΔT\) produced by the combustion. This is gotten by dividing the calculated heat of combustion by the heat capacity of the calorimeter assembly. Use the formula \(\Delta T = \frac{|q_p|}{C} = \frac{|-16.50\, \mathrm{kJ}|}{3.87\, \mathrm{kJ/ ^{\circ}C}}\). This will give you the temperature change of the bomb calorimeter.

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Most popular questions from this chapter

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$$C_{P, \mathrm{m}}=28.58+3.77 \times 10^{-3} T-\frac{0.5 \times 10^{5}}{T^{2}} \mathrm{JK}^{-1} \mathrm{mol}^{-1}$$

The standard molar heats of combustion of C(graphite) and \(\mathrm{CO}(\mathrm{g})\) are -393.5 and \(-283 \mathrm{kJ} / \mathrm{mol}\) respectively. Use those data and that for the following reaction $$\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{COCl}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-108 \mathrm{kJ}$$ to calculate the standard molar enthalpy of formation of \(\mathrm{COCl}_{2}(\mathrm{g})\).

A coffee-cup calorimeter contains \(100.0 \mathrm{mL}\) of \(0.300 \mathrm{M}\) HCl at \(20.3^{\circ} \mathrm{C}\). When \(1.82 \mathrm{g} \mathrm{Zn}(\mathrm{s})\) is added, the temperature rises to \(30.5^{\circ} \mathrm{C}\). What is the heat of reaction per mol Zn? Make the same assumptions as in Example \(7-4,\) and also assume that there is no heat lost with the \(\mathrm{H}_{2}(\mathrm{g})\) that escapes. $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})$$

Is it possible for a chemical reaction to have \(\Delta U<0\) and \(\Delta H>0 ?\) Explain.

The following substances undergo complete combustion in a bomb calorimeter. The calorimeter assembly has a heat capacity of \(5.136 \mathrm{kJ} /^{\circ} \mathrm{C} .\) In each case, what is the final temperature if the initial water temperature is \(22.43^{\circ} \mathrm{C} ?\) \(\begin{array}{lllll}\text { (a) } 0.3268 & \text { g caffeine, } & \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{N}_{4} & \text { (heat of }\end{array}\) combustion \(=-1014.2 \mathrm{kcal} / \mathrm{mol} \text { caffeine })\) (b) \(1.35 \mathrm{mL}\) of methyl ethyl ketone, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}(1)\) \(d=0.805 \mathrm{g} / \mathrm{mL}\) (heat of combustion \(=-2444 \mathrm{kJ} / \mathrm{mol}\) methyl ethyl ketone).
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