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Chapter 6: Problem 20

Explain why \(\Delta H\) is obtained directly from coffee-cup calorimeters, whereas \(\Delta E\) is obtained directly from bomb calorimeters.

Short Answer

Expert verified
Coffee-cup calorimeters measure heat changes at constant pressure (isobaric conditions), which is directly related to the enthalpy change (∆H) of the reaction. On the other hand, bomb calorimeters measure heat changes at constant volume (isochoric conditions), which is directly related to the internal energy change (∆E) of the reaction. Therefore, ∆H is obtained directly from coffee-cup calorimeters, whereas ∆E is obtained directly from bomb calorimeters.

Step by step solution

01

Understanding Coffee-cup Calorimeters

Coffee-cup calorimeters are simple, open-system calorimeters that are well insulated to minimize heat exchange with the surroundings. As a result, these calorimeters measure heat changes at constant pressure (isobaric conditions). Hence, the heat change measured by a coffee-cup calorimeter is directly related to the enthalpy change (∆H) of the reaction.
02

Understanding Bomb Calorimeters

Bomb calorimeters, on the other hand, are closed-system calorimeters that consist of a sealed metal container called a "bomb" in which the reaction takes place, submerged in a known amount of water. The pressure inside the bomb is allowed to increase as the reaction proceeds, but the volume is kept constant. As a result, bomb calorimeters measure heat changes at constant volume (isochoric conditions). The heat change measured at constant volume is directly related to the internal energy change (∆E) of the reaction.
03

Comparing Coffee-cup and Bomb Calorimeters

To summarize, coffee-cup calorimeters measure heat changes at constant pressure, while bomb calorimeters measure heat changes at constant volume. Since enthalpy changes (∆H) occur at constant pressure and internal energy changes (∆E) occur at constant volume, we can conclude that ∆H is obtained directly from coffee-cup calorimeters and ∆E is obtained directly from bomb calorimeters.

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Most popular questions from this chapter

The complete combustion of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g),\) produces 1300 . kJ of energy per mole of acetylene consumed. How many grams of acetylene must be burned to produce enough heat to raise the temperature of 1.00 gal water by \(10.0^{\circ} \mathrm{C}\) if the process is 80.0\(\%\) efficient? Assume the density of water is 1.00 \(\mathrm{g} / \mathrm{cm}^{3}\)

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Consider the following reaction: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{kJ} $$ a. How much heat is evolved for the production of 1.00 mole of $\mathrm{H}_{2} \mathrm{O}(l) ?$ b. How much heat is evolved when 4.03 g hydrogen are reacted with excess oxygen? c. How much heat is evolved when 186 \(\mathrm{g}\) oxygen are reacted with excess hydrogen? d. The total volume of hydrogen gas needed to fill the Hindenburg was $2.0 \times 10^{8} \mathrm{L}\( at 1.0 atm and \)25^{\circ} \mathrm{C} .$ How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?

Consider the reaction $$ 2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ $$ \Delta H=-118 \mathrm{kJ} $$ Calculate the heat when 100.0 \(\mathrm{mL}\) of 0.500\(M \mathrm{HCl}\) is mixed with 300.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{Ba}(\mathrm{OH})_{2}\) . Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of 400.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) , calculate the final temperature of the mixture.

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area \((1 \text { watt }=1 \mathrm{Js} \text { ). The plants in an }\) agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{m}^{2}\right) .\) Assuming that sucrose is produced by the reaction $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+& 12 \mathrm{O}_{2}(g) \\ & \Delta H=5640 \mathrm{kJ} \end{aligned} $$ calculate the percentage of sunlight used to produce the sucrose-that is, determine the efficiency of photosynthesis.
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