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Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{~kJ}\) and \(w=-840 \mathrm{~J}\). (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.

Short Answer

Expert verified
(a) The process has a ∆E of -0.077 kJ, and since q is positive (0.763 kJ), it is an endothermic process. (b) The process has a ∆E of -22.1 kJ, and since q is negative (-66.1 kJ), it is an exothermic process.

Step by step solution

01

Calculate ∆E

Use the equation ∆E = q + w. For this case, q = 0.763 kJ and w = -840 J. First, let's convert w to kJ: w = -840 J × (1 kJ / 1000 J) = -0.840 kJ. Now we can calculate ∆E: ∆E = 0.763 kJ - 0.840 kJ = -0.077 kJ.
02

Determine if endothermic or exothermic

Since q > 0 in this case (0.763 kJ), the process is endothermic according to equation 2. Case (b):
03

Calculate q and w

We are given that the system releases 66.1 kJ of heat (q = -66.1 kJ, since heat is released), and the surroundings do 44.0 kJ of work on the system (w = +44.0 kJ, since work is being done on the system).
04

Calculate ∆E

Use the equation ∆E = q + w. In this case, ∆E = -66.1 kJ + 44.0 kJ = -22.1 kJ.
05

Determine if endothermic or exothermic

Since q < 0 in this case (-66.1 kJ), the process is exothermic according to equation 3.

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Key Concepts

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

Endothermic process
An endothermic process is one in which a system absorbs heat from its surroundings. This means that energy is required to make the process happen. The absorbed energy usually comes in the form of heat. A common way to figure out if a process is endothermic is to look at the heat exchanged, denoted as \( q \).
- If \( q > 0 \), this indicates that the system takes in heat, marking the process as endothermic.- Such processes often involve melting, boiling, or subliming of substances.
For instance, when ice melts to form water, it requires energy absorbed from the surroundings, making it endothermic. In the given exercise, case (a) with \( q = 0.763 \text{ kJ} \) shows the system absorbs heat, confirming an endothermic process.
Exothermic process
Exothermic processes are the opposite of endothermic processes. Here, a system releases heat to its surroundings. This results in a decrease in the energy of the system while increasing the energy of the surroundings. We can determine if a process is exothermic by examining the sign of \( q \).
- If \( q < 0 \), it means the system is giving off heat, marking it as exothermic.- Common examples include combustion, condensation, and freezing.
Exothermic processes usually make the surroundings warmer as they release energy. For instance, when water vapor condenses into liquid, it releases energy into the environment. In the exercise, case (b) releases \(66.1 \text{ kJ}\) of heat to its surroundings. This negative \( q \) value identifies it as exothermic.
Energy change (ΔE)
Energy change, denoted by \( \Delta E \), represents the total change in a system’s internal energy. It combines the heat exchange \( q \), and the work done \( w \) using:\[\Delta E = q + w\]
- Work typically relates to expansion or contraction of gas, which is expressed as work done by or on the system.- Converting units may be necessary, for instance, if work \( w \) is given in joules, and you need to calculate \( \Delta E \) in kilojoules.
For case (a), converting \( w = -840 \text{ J} \) to \( \text{kJ} \) helps solve \( \Delta E \) as reflectively detailed in the calculation steps. Both case (a) and (b) emphasize how changes in \( q \) and \( w \) affect the energy state of the system. In case (b), a large negative heat exchange leads to an overall negative \( \Delta E \), indicative of net energy loss, a characteristic of exothermic reactions.

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Most popular questions from this chapter

At the end of 2012 , global population was about \(7.0\) billion people. What mass of glucose in kg would be needed to provide \(1500 \mathrm{cal} /\) person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) according to the following thermochemical equation: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+& 6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-2803 \mathrm{~kJ} \end{aligned} $$

5.104 We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) $$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$ \begin{array}{ll} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-890.3 \mathrm{~kJ} \\ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) & \Delta H^{\circ}=-136.3 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-571.6 \mathrm{~kJ} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-3120.8 \mathrm{~kJ} \end{array} $$

Complete combustion of \(1 \mathrm{~mol}\) of acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\) liberates \(1790 \mathrm{~kJ}\) : $$ \begin{aligned} \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+& 3 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-1790 \mathrm{~kJ} \end{aligned} $$ Using this information together with the standard enthalpies of formation of \(\mathrm{O}_{2}(g), \mathrm{CO}_{2}(g)\), and \(\mathrm{H}_{2} \mathrm{O}(l)\) from Appendix \(\mathrm{C}\), calculate the standard enthalpy of formation of acetone.

How much work (in J) is involved in a chemical reaction if the volume decreases from \(5.00\) to \(1.26 \mathrm{~L}\) against a constant pressure of \(0.857 \mathrm{~atm}\) ?

Suppose that the gas-phase reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow\) \(2 \mathrm{NO}_{2}(g)\) were carried out in a constant-volume container at constant temperature. (a) Would the measured heat change represent \(\Delta H\) or \(\Delta E\) ? (b) If there is a difference, which quantity is larger for this reaction? (c) Explain your answer to part (b).

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