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Chapter 17: Problem 36

At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\) c. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

Short Answer

Expert verified
The processes will be spontaneous under the following conditions: a. \(T > 300\mathrm{~K}\) b. \(T < 300\mathrm{~K}\) c. Never spontaneous d. Always spontaneous

Step by step solution

01

Case a: 螖H = -18 kJ, 螖S = -60 J/K

Let's solve for T when \( 螖G = 螖H - T螖S < 0 \): \[ -18\mathrm{~kJ} - T(-60 . \mathrm{J} / \mathrm{K}) < 0 \] First, convert 鈭咹 to J: \[ -18000\mathrm{~J} + 60T < 0 \] Solve for T: \[ T > \frac{18000}{60} \Rightarrow T > 300\mathrm{~K} \] So, in this case, the process is spontaneous when the temperature is greater than 300 K.
02

Case b: 螖H = +18 kJ, 螖S = +60 J/K

Let's solve for T when \( 螖G = 螖H - T螖S < 0 \): \[ 18\mathrm{~kJ} - T(60 . \mathrm{J} / \mathrm{K}) < 0 \] Convert 鈭咹 to J: \[ 18000\mathrm{~J} - 60T < 0 \] Solve for T: \[ T < \frac{18000}{60} \Rightarrow T < 300\mathrm{~K} \] In this case, the process is spontaneous when the temperature is less than 300 K.
03

Case c: 螖H = +18 kJ, 螖S = -60 J/K

Let's solve for T when \( 螖G = 螖H - T螖S < 0 \): \[ 18\mathrm{~kJ} - T(-60 . \mathrm{J} / \mathrm{K}) < 0 \] Convert 鈭咹 to J: \[ 18000\mathrm{~J} + 60T < 0 \] Solve for T: \[ T < -\frac{18000}{60} \Rightarrow T < -300\mathrm{~K} \] However, temperature cannot be negative. Therefore, the process is never spontaneous in this case.
04

Case d: 螖H = -18 kJ, 螖S = +60 J/K

Let's solve for T when \( 螖G = 螖H - T螖S < 0 \): \[ -18\mathrm{~kJ} - T(60 . \mathrm{J} / \mathrm{K}) < 0 \] Convert 鈭咹 to J: \[ -18000\mathrm{~J} - 60T < 0 \] Solve for T: \[ T > -\frac{-18000}{60} \Rightarrow T > -300\mathrm{~K} \] Since absolute temperature is always positive, the process is always spontaneous in this case.

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Key Concepts

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

Thermodynamics
Thermodynamics is the study of energy transformations in physical systems, and it plays a crucial role in understanding chemical reactions and processes. At the core of thermodynamics is the concept of energy conservation, which states that energy cannot be created or destroyed, only transformed from one form to another.
  • Gibbs Free Energy (螖G) is a crucial thermodynamic quantity that helps predict whether a reaction is spontaneous under constant temperature and pressure.
  • The formula for Gibbs Free Energy is given by: \( \Delta G = \Delta H - T\Delta S \), where:
    • \( \Delta H \) is the change in enthalpy (heat content).
    • \( \Delta S \) is the change in entropy (degree of disorder).
    • \( T \) is the absolute temperature in Kelvin.
By calculating Gibbs Free Energy, we can determine whether a chemical process will occur spontaneously.
Enthalpy
Enthalpy is a measure of the total heat content in a system and is represented by \( \Delta H \). It reflects the energy absorbed or released during a chemical reaction or phase change at constant pressure.
  • A negative \( \Delta H \) value indicates that the process is exothermic, meaning heat is released to the surroundings.
  • A positive \( \Delta H \) value, on the other hand, indicates an endothermic process, where heat is absorbed from the surroundings.
Enthalpy changes are an essential factor in determining the spontaneity of a reaction. They show how the system's heat content varies, directly affecting the Gibbs Free Energy.
Entropy
Entropy, symbolized as \( \Delta S \), is a measure of the level of disorder or randomness in a system. Globally, it illustrates how energy disperses, with higher entropy indicating a greater degree of disorder.
  • If \( \Delta S \) is positive, the disorder increases during the process, making it more likely to be spontaneous.
  • Conversely, a negative \( \Delta S \) suggests a decrease in disorder, which tends to be less favorable for spontaneity.
Entropy is central to the second law of thermodynamics which states that the total entropy of a closed system always increases over time.
Spontaneity
Spontaneity in a chemical reaction refers to the ability of a process to occur naturally without external intervention. Whether a reaction is spontaneous at a given temperature can often be determined by calculating Gibbs Free Energy (\( \Delta G \)).
  • When \( \Delta G < 0 \), the process is spontaneous.
  • If \( \Delta G > 0 \), the process is non-spontaneous and requires energy input to proceed.
  • At \( \Delta G = 0 \), the system is at equilibrium, with no net change occurring.
Understanding spontaneity helps in predicting which chemical reactions will occur under certain conditions.
Temperature Dependence
Temperature is a key factor influencing the spontaneity of reactions. As seen in Gibbs Free Energy's equation \( \Delta G = \Delta H - T\Delta S \), temperature (
  • When \( \Delta H \) is negative and \( \Delta S \) is positive, the process is always spontaneous at all temperatures.
  • If both \( \Delta H \) and \( \Delta S \) are positive, the reaction is spontaneous at high temperatures where the \( T\Delta S \) term can outweigh \( \Delta H \).
  • Conversely, if both are negative, lower temperatures favor spontaneity.
  • Temperature affects how entropy and enthalpy interact to influence \( \Delta G \), mapping the range of conditions under which a process becomes favorable or unfavorable.
Assessing temperature dependence is essential to predict and manipulate chemical reactions in practical applications.

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Most popular questions from this chapter

As \(\mathrm{O}_{2}(l)\) is cooled at \(1 \mathrm{~atm}\), it freezes at \(54.5 \mathrm{~K}\) to form solid \(\mathrm{I}\). At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the \(\mathrm{I} \rightarrow\) II phase transition is \(-743.1 \mathrm{~J} / \mathrm{mol}\), and \(\Delta S\) for the same transition is \(-17.0 \mathrm{~J} / \mathrm{K} \cdot\) mol. At what temperature are solids I and II in equilibrium?

Given the following data: $$ \begin{aligned} 2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow & 12 \mathrm{CO}_{2}(g)+& 6 \mathrm{H}_{2} \mathrm{O}(l) \\ & \Delta G^{\circ}=-6399 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta G^{\circ}\) for the reaction $$ 6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l) $$

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$ \begin{array}{cl} \mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array} $$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ \mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2} $$

Gas \(\mathrm{A}_{2}\) reacts with gas \(\mathrm{B}_{2}\) to form gas \(\mathrm{AB}\) at a constant temperature. The bond energy of \(\mathrm{AB}\) is much greater than that of either reactant. What can be said about the sign of \(\Delta H ? \Delta S_{\text {surr }}\) ? \(\Delta S\) ? Explain how potential energy changes for this process. Explain how random kinetic energy changes during the process.

Some nonelectrolyte solute (molar mass \(=142 \mathrm{~g} / \mathrm{mol}\) ) was dissolved in \(150 . \mathrm{mL}\) of a solvent \(\left(\right.\) density \(\left.=0.879 \mathrm{~g} / \mathrm{cm}^{3}\right)\). The elevated boiling point of the solution was \(355.4 \mathrm{~K}\). What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is \(33.90 \mathrm{~kJ} / \mathrm{mol}\), the entropy of vaporization is \(95.95 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\), and the boiling-point elevation constant is \(2.5 \mathrm{~K} \cdot \mathrm{kg} / \mathrm{mol}\).
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