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Chapter 7: Problem 36

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).

Short Answer

Expert verified
The final temperature of the water after the combustion of caffeine is 22.747°C, and for methyl ethyl ketone it is 29.575°C.

Step by step solution

01

Calculate number of moles of the first substance

Calculate the number of moles (n₁) for caffeine as the molecular weight of caffeine is 194.19 g/mol.\n\n\[ n₁ = 0.3268/194.19 = 0.001682 mol \]
02

Calculate change in temperature for the first substance

Next, the change in temperature (∆t₁) due to combustion of caffeine can be calculated using the relation ∆t = - ( ∆Hc ⋅ n ) / C.\n\n\[ ∆t₁ = - ( -1014.2 × 0.001682 ) / 5.136 = 0.317°C \]
03

Final temperature for the first substance

Now we can find the final temperature of the water (T₁) for caffeine combustion, by adding the initial temperature to ∆t₁.\n\n\[ T₁ = 22.43 + 0.317 = 22.747°C \]
04

Calculate number of moles of the second substance

Now, calculate the number of moles (n₂) for methyl ethyl ketone (C₄H₈O). The molecular weight of C₄H₈O is 72.105 g/mol, and the given volume is 1.35 mL with a density of 0.805 g/mL, from which we get the mass.\n\n\[ m = 1.35 × 0.805 = 1.08675 g \]\n\n\[ n₂ = 1.08675/72.105 = 0.015064 mol \]
05

Calculate change in temperature for the second substance

Now, the change in temperature (∆t₂) due to combustion of methyl ethyl ketone can be calculated in a similar manner as in Step 2. However, we use 2444 kJ/mol instead of kcal/mol and also insert the heat of combustion with a negative value.\n\n\[ ∆t₂ = - ( -2444 × 0.015064 ) / 5.136 = 7.145°C \]
06

Final temperature for the second substance

Finally, the final temperature of the water upon combustion of methyl ethyl ketone (T₂) can be calculated by adding the initial temperature to ∆t₂.\n\n\[ T₂ = 22.43 + 7.145 = 29.575°C \]

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Most popular questions from this chapter

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction \(\mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) given that $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+4 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow & 3 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ && \Delta H^{\circ}=-1937 \mathrm{kJ} \end{aligned}$$ $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.1 \mathrm{kJ} \end{array}$$

James Joule published his definitive work related to the first law of thermodynamics in \(1850 .\) He stated that "the quantity of heat capable of increasing the temperature of one pound of water by \(1^{\circ} \mathrm{F}\) requires for its evolution the expenditure of a mechanical force represented by the fall of 772 lb through the space of one foot." Validate this statement by relating it to information given in this text.

Under the entry \(\mathrm{H}_{2} \mathrm{SO}_{4},\) a reference source lists many values for the standard enthalpy of formation. For example, for pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1), \Delta H_{\mathrm{f}}^{\circ}=-814.0 \mathrm{kJ} / \mathrm{mol}\) for a solution with \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(-841.8 ;\) with \(10 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-880.5 ;\) with \(50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-886.8 ;\) with \(100 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-887.7 ;\) with \(500 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-890.5 ;\) with \(1000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-892.3 ;\) with \(10,000 \mathrm{mol}\) \(\mathrm{H}_{2} \mathrm{O},-900.8 ;\) and with \(100,000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-907.3\) (a) Explain why these values are not all the same. (b) The value of \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right]\) in an infinitely dilute solution is \(-909.3 \mathrm{kJ} / \mathrm{mol} .\) What data from this chapter can you cite to confirm this value? Explain. (c) If \(500.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is prepared from pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1),\) what is the approximate change in temperature that should be observed? Assume that the \(\mathrm{H}_{2} \mathrm{SO}_{4}(1)\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are at the same temperature initially and that the specific heat of the \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is about \(4.2 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

The combustion of methane gas, the principal constituent of natural gas, is represented by the equation $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ} &=-890.3 \mathrm{kJ} \end{aligned}$$ (a) What mass of methane, in kilograms, must be burned to liberate \(2.80 \times 10^{7} \mathrm{kJ}\) of heat? (b) What quantity of heat, in kilojoules, is liberated in the complete combustion of \(1.65 \times 10^{4} \mathrm{L}\) of \(\mathrm{CH}_{4}(\mathrm{g})\) measured at \(18.6^{\circ} \mathrm{C}\) and \(768 \mathrm{mmHg} ?\) (c) If the quantity of heat calculated in part (b) could be transferred with \(100 \%\) efficiency to water, what volume of water, in liters, could be heated from 8.8 to \(60.0^{\circ} \mathrm{C}\) as a result?

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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