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Chapter 17: Problem 8

Is \(\Delta S_{\text {surr }}\) favorable or unfavorable for exothermic reactions? Endothermic reactions? Explain.

Short Answer

Expert verified
For exothermic reactions, the change in entropy of the surroundings (\(\Delta S_{surr}\)) is favorable, as it becomes more disordered (positive \(\Delta S_{surr}\)). In contrast, for endothermic reactions, the change in entropy of the surroundings is unfavorable, as it becomes more ordered (negative \(\Delta S_{surr}\)). This is due to the heat transfer from the system during these reactions, with exothermic reactions releasing heat and endothermic reactions absorbing heat.

Step by step solution

01

Define exothermic and endothermic reactions

An exothermic reaction is a chemical reaction that releases energy in the form of heat. In contrast, an endothermic reaction is a chemical reaction that absorbs energy from the surroundings in the form of heat.
02

Introduce the concept of entropy

Entropy (S) is a thermodynamic quantity that represents the degree of disorder or randomness in a system. The change in entropy, denoted as 螖S, is positive when the system becomes more disordered and negative when the system becomes more ordered.
03

Relate heat and entropy

The relationship between heat transfer (q) and the change in entropy (螖S) of the surroundings can be expressed by the following equation: \[ \Delta S_{surr} = -\frac{q_{sys}}{T} \] where 螖Ssurr is the change in entropy of the surroundings, qsys is the heat generated by the system (positive for endothermic reactions, negative for exothermic reactions), and T is the temperature of the surroundings in Kelvin.
04

Determine the change in entropy for exothermic reactions

For exothermic reactions, the heat transfer (qsys) to the surroundings is negative (i.e., heat is released to the surroundings). Therefore, the equation for the change in entropy becomes: \[ \Delta S_{surr} = -\frac{-q_{sys}}{T} \] Since both qsys and T are positive, the 螖Ssurr must also be positive. This indicates that the surroundings become more disordered, which is considered favorable for natural processes.
05

Determine the change in entropy for endothermic reactions

For endothermic reactions, the heat transfer (qsys) from the surroundings is positive (i.e., heat is absorbed by the system). Therefore, the equation for the change in entropy becomes: \[ \Delta S_{surr} = -\frac{q_{sys}}{T} \] Since both qsys and T are positive, the 螖Ssurr must be negative. This indicates that the surroundings become more ordered, which is considered unfavorable for natural processes. In conclusion, the change in entropy of the surroundings (螖Ssurr) is favorable for exothermic reactions (positive 螖Ssurr) and unfavorable for endothermic reactions (negative 螖Ssurr).

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Most popular questions from this chapter

For mercury, the enthalpy of vaporization is \(58.51 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of vaporization is \(92.92 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). What is the normal boiling point of mercury?

Calculate \(\Delta S_{\text {surr }}\) for the following reactions at \(25^{\circ} \mathrm{C}\) and 1 atm. a. \(\mathrm{C}_{3} \mathrm{H}_{5}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(t) \Delta H^{\circ}=-2221 \mathrm{~kJ}\) b. \(2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\) \(\Delta H^{\circ}=112 \mathrm{~kJ}\)

Consider the system $$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$ at \(25^{\circ} \mathrm{C}\). a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}\), cal- culate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if \(1.00 \mathrm{~mol} \mathrm{~A}(\mathrm{~g})\) at \(1.00\) atm and \(1.00 \mathrm{~mol} \mathrm{~B}(g)\) at \(1.00 \mathrm{~atm}\) are mixed at \(25^{\circ} \mathrm{C}\). c. Show by calculations that \(\Delta G=0\) at equilibrium.

Using data from Appendix 4, calculate \(\Delta H^{\circ}, \Delta G^{\circ}\), and \(K\) (at 298 K) for the production of ozone from oxygen: $$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$ At \(30 \mathrm{~km}\) above the surface of the earth, the temperature is about 230\. \(\mathrm{K}\) and the partial pressure of oxygen is about \(1.0 \times 10^{-3}\) atm. Estimate the partial pressure of ozone in equilibrium with oxygen at \(30 \mathrm{~km}\) above the earth's surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.

Using data from Appendix 4, calculate \(\Delta G\) for the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{~S}_{\text {mombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ for the following conditions at \(25^{\circ} \mathrm{C}\) : $$\begin{array}{l}P_{\mathrm{H}_{2} \mathrm{~S}}=1.0 \times 10^{-4} \mathrm{~atm} \\\P_{\mathrm{SO}_{2}}=1.0 \times 10^{-2} \mathrm{~atm} \\ P_{\mathrm{H}_{2} \mathrm{O}}=3.0 \times 10^{-2} \mathrm{~atm}\end{array}$$
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