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Chapter 5: Problem 27

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).

Short Answer

Expert verified
For the given cases, we have calculated the following values for \(\Delta E\): - Case (a): \(\Delta E = -77 \mathrm{~J}\), indicating an exothermic process. - Case (b): \(\Delta E = -22100 \mathrm{~J}\), indicating an exothermic process. - Case (c): \(\Delta E = 7250 \mathrm{~J}\), indicating an endothermic process.

Step by step solution

01

Case (a)

Calculate \(\Delta E\) and determine if the process is endothermic or exothermic. We have q = 0.763 kJ and w = -840 J. First, we need to convert q into joules: 0.763 kJ * 1000 J/kJ = 763 J Now we can use the formula: \[\Delta E = q + w = 763 \mathrm{~J} + (-840 \mathrm{~J}) = -77 \mathrm{~J}\] Since \(\Delta E\) is negative, the process is exothermic, which means it releases heat.
02

Case (b)

First, let's establish the signs of q and w. Since the system releases heat, q is negative. Since the surroundings do work on the system, w is positive. Now, convert the given kJ to J: q = -66.1 kJ * 1000 J/kJ = -66100 J w = 44.0 kJ * 1000 J/kJ = 44000 J Next, we'll calculate \(\Delta E\): \[\Delta E = q + w = (-66100 \mathrm{~J}) + (44000 \mathrm{~J}) = -22100 \mathrm{~J}\] With a negative \(\Delta E\), the process is exothermic, meaning it releases heat.
03

Case (c)

Since the volume remains constant and only \(P-V\) work can be done, there is no work being done in this case (w = 0). The system absorbs heat from the surrounding so q is positive: q = 7.25 kJ * 1000 J/kJ = 7250 J Now compute \(\Delta E\): \[\Delta E = q + w = 7250 \mathrm{~J} + 0 = 7250 \mathrm{~J}\] Since \(\Delta E\) is positive, the process is endothermic, which means it absorbs heat.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Endothermic Processes
Endothermic processes are fascinating. In these processes, a system absorbs energy from its surroundings. This energy is usually in the form of heat.In case (c) from our original exercise, we encountered an endothermic reaction. Here, the system absorbed heat, with a calculated energy change (\(\Delta E = 7250 \mathrm{~J}\)). This positive energy change indicates that the system gained energy. With endothermic reactions, temperature changes can often feel cold to the touch, as heat is absorbed rather than released.Let's briefly explore some common endothermic processes:
  • Melting ice.
  • Evaporating water.
  • Photosynthesis in plants.
In many of these processes, the system (be it ice, water, or a plant) requires energy to move to a higher energy state. Endothermic reactions are essential in both natural cycles and industrial processes.
Exothermic Processes
Exothermic processes release energy to the surroundings. This type of energy release typically involves heat, resulting in an increase in temperature around the system.Examining cases (a) and (b) from our exercise helps emphasize this concept. Both instances resulted in a negative \(\Delta E\), indicating a release of energy:- Case (a) had \(\Delta E = -77 \mathrm{~J}\).- Case (b) had \(\Delta E = -22100 \mathrm{~J}\).Thermochemically, this loss of energy is felt as heat generation. Exothermic reactions are everywhere:
  • Combustion, like burning wood or gasoline.
  • Respiration in living cells.
  • Mixing strong acids and bases.
In these reactions, energy is released when bonds form in reaction products. Exothermic reactions are vital in many applications, such as engine fuels and metabolic processes in organisms.
Energy Change Calculations
Calculating energy changes requires a clear understanding of the relationship between heat (\(q\)) and work (\(w\)). This relationship is expressed as:\[ \Delta E = q + w \]Let's detail this formula using our exercise:
  • Case (a): Converted heat and work gave us \(\Delta E = -77 \mathrm{~J}\).
  • Case (b): Converted measurements resulted in \(\Delta E = -22100 \mathrm{~J}\).
  • Case (c): With a constant volume, w = 0, leaving \(\Delta E = 7250 \mathrm{~J}\) solely from the absorbed heat.
Notice the need to convert kilojoules to joules. This step ensures consistency in units, making the calculations yielding reliable results easier. Understanding \(\Delta E\) helps predict whether a reaction absorbs or releases energy. This understanding is foundational in thermochemistry, directly impacting how we control reactions in lab and industrial settings.Precision in energy change calculations not only determines heating and cooling requirements but also ensures efficiency in energy usage across various processes.

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Most popular questions from this chapter

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\(]\)

Consider two solutions, the first being \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{CuSO}_{4}\) and the second \(50.0 \mathrm{~mL}\) of \(2.00 \mathrm{MKOH}\). When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) to \(27.7^{\circ} \mathrm{C}\). (a) Before mixing, how many grams of Cu are present in the solution of \(\mathrm{CuSO}_{4} ?\) (b) Predict the identity of the precipitate in the reaction. (c) Write complete and net ionic equations for the reaction that occurs when the two solutions are mixed. (d) From the calorimetric data, calculate \(\Delta H\) for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 \(\mathrm{mL},\) and that the specific heat and density of the solution after mixing are the same as that of pure water.

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l):\) $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H=&-726.5 \mathrm{~kJ} \end{aligned} $$ (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(g)\) instead of \(\mathrm{H}_{2} \mathrm{O}(l)\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

A sample of a hydrocarbon is combusted completely in \(\mathrm{O}_{2}(g)\) to produce \(21.83 \mathrm{~g} \mathrm{CO}_{2}(g), 4.47 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}(g),\) and \(311 \mathrm{~kJ}\) of heat. (a) What is the mass of the hydrocarbon sample that was combusted? (b) What is the empirical formula of the hydrocarbon? (c) Calculate the value of \(\Delta H_{f}^{\circ}\) per empirical-formula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.

Consider the following reaction: \(2 \mathrm{CH}_{3} \mathrm{OH}(g) \longrightarrow 2 \mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \quad \Delta H=+252.8 \mathrm{~kJ}\) (a) Is this reaction exothermic or endothermic? (b) Calculate the amount of heat transferred when \(24.0 \mathrm{~g}\) of \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is decomposed by this reaction at constant pressure. (c) For a given sample of \(\mathrm{CH}_{3} \mathrm{OH},\) the enthalpy change during the reaction is \(82.1 \mathrm{~kJ}\). How many grams of methane gas are produced? (d) How many kilojoules of heat are released when \(38.5 \mathrm{~g}\) of \(\mathrm{CH}_{4}(g)\) reacts completely with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{CH}_{3} \mathrm{OH}(g)\) at constant pressure?
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