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Chapter 5: Problem 60

Under constant-volume conditions, the heat of combustion of benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is \(26.38 \mathrm{~kJ} / \mathrm{g}\). A 2.760 -g sample of benzoic acid is burned in a bomb calorimeter. The temperature of the calorimeter increases from \(21.60^{\circ} \mathrm{C}\) to \(29.93^{\circ} \mathrm{C}\). (a) What is the total heat capacity of the calorimeter? (b) \(\mathrm{A}\) 1.440-g sample of a new organic substance is combusted in the same calorimeter. The temperature of the calorimeter increases from \(22.14^{\circ} \mathrm{C}\) to \(27.09^{\circ} \mathrm{C}\). What is the heat of combustion per gram of the new substance? (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

Short Answer

Expert verified
In summary: (a) The total heat capacity of the calorimeter is 8.74 kJ/°C. (b) The heat of combustion per gram of the new substance is 30.04 kJ/g. (c) Losing a portion of water in the calorimeter could decrease its overall heat capacity; however, the effect might not be significant if the change is small. In case of substantial water loss, recalibration of the calorimeter would be necessary to determine the new heat capacity.

Step by step solution

01

Part (a): Heat Capacity of the Calorimeter

First, we need to calculate the total heat released by the combustion of benzoic acid (ΔH_benzoic). We know that the heat of combustion of benzoic acid is: ΔH_benzoic = 26.38 kJ/g Given that we have 2.760 g of benzoic acid, the total heat released during its combustion is: ΔH_total = ΔH_benzoic × mass ΔH_total = 26.38 kJ/g × 2.760 g = 72.8072 kJ Now, we know the temperature change of the calorimeter during the combustion of benzoic acid: ΔT_calorimeter = T_final - T_initial ΔT_calorimeter = 29.93°C - 21.60°C = 8.33°C We can now find the total heat capacity (C_calorimeter) of the calorimeter: C_calorimeter = ΔH_total / ΔT_calorimeter C_calorimeter = 72.8072 kJ / 8.33°C ≈ 8.74 kJ/°C
02

Part (b): Heat of Combustion of the New Substance

To determine the heat of combustion per gram of the new substance, we first need to calculate the total heat released during its combustion: The temperature change of the calorimeter during the combustion of the new substance is: ΔT_calorimeter_new = T_final - T_initial ΔT_calorimeter_new = 27.09°C - 22.14°C = 4.95°C Now, we can use the heat capacity of the calorimeter to find the total heat released by the new substance: ΔH_total_new = C_calorimeter × ΔT_calorimeter_new ΔH_total_new = 8.74 kJ/°C × 4.95°C ≈ 43.26 kJ We are also given that the mass of the new substance is 1.440 g. We can now calculate the heat of combustion per gram of the new substance (ΔH_combustion_new): ΔH_combustion_new = ΔH_total_new / mass_new ΔH_combustion_new = 43.26 kJ / 1.440 g ≈ 30.04 kJ/g
03

Part (c): Effect of water loss on calorimeter heat capacity

If a portion of the water in the calorimeter is lost, it could decrease the overall heat capacity of the calorimeter because the water generally has a higher specific heat capacity than the other components of the calorimeter. However, if the change is small, it might not have a significant effect on the results of the experiment. If the water loss is substantial, it might be necessary to recalibrate the calorimeter to find the new heat capacity before performing further experiments.

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Most popular questions from this chapter

A pound of plain \(\mathrm{M\&M}\) candies contains \(96 \mathrm{~g}\) fat, \(320 \mathrm{~g}\) carbohydrate, and 21 g protein. What is the fuel value in \(\mathrm{kJ}\) in a \(42-\mathrm{g}\) (about 1.5 oz ) serving? How many Calories does it provide?

The hydrocarbons acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) have the same empirical formula. Benzene is an "aromatic" hydrocarbon, one that is unusually stable because of its structure. (a) By using data in Appendix \(\mathrm{C}\), determine the standard enthalpy change for the reaction $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$.

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{CO}_{2}(s)\) or \(1 \mathrm{~mol}\) \(\mathrm{CO}_{2}(g)\) at the same temperature, (b) 2 mol of hydrogen atoms or \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2},\) (c) \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) and \(0.5 \mathrm{~mol} \mathrm{O}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(g)\) at \(25^{\circ} \mathrm{C},\) (d) \(1 \mathrm{~mol} \mathrm{~N}_{2}(g)\) at \(100{ }^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol}\) \(\mathrm{N}_{2}(g)\) at \(300^{\circ} \mathrm{C}\)

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{~kJ}\) and \(w=-840 \mathrm{~J} ;(\mathbf{b})\) a system releases \(66.1 \mathrm{~kJ}\) of heat to its surroundings while the surroundings do \(44.0 \mathrm{~kJ}\) of work on the system; (c) the system absorbs \(7.25 \mathrm{~kJ}\) of heat from the surroundings while its volume remains constant (assume that only \(P-V\) work can be done).

You may have noticed that when you compress the air in a bicycle pump, the body of the pump gets warmer. (a) Assuming the pump and the air in it comprise the system, what is the sign of \(w\) when you compress the air? (b) What is the sign of \(q\) for this process? (c) Based on your answers to parts (a) and (b), can you determine the sign of \(\Delta E\) for compressing the air in the pump? If not, what would you expect for the sign of \(\Delta E\) ? What is your reasoning? [Section 5.2\(]\)
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