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Chapter 19: Problem 56

A certain reaction has \(\Delta H^{\circ}=+20.0 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+100.0 \mathrm{~J} / \mathrm{K} .\) (a) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K} .(\mathbf{d})\) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Short Answer

Expert verified
(a) The reaction leads to an increase in the randomness or disorder of the system, as 螖S鈦� is positive. (b) The reaction leads to a decrease in the randomness or disorder of the surroundings, as 螖H鈦� is positive (endothermic reaction). (c) At 298 K, 螖G鈦� = -9,800 J. (d) The reaction is spontaneous at 298 K under standard conditions, as 螖G鈦� is negative.

Step by step solution

01

Analyzing the change in randomness or disorder of the system

We are given the entropy change 螖S鈦� as +100.0 J/K. Since the value is positive, it implies that the reaction leads to an increase in the randomness or disorder of the system.
02

Analyzing the change in randomness or disorder of the surroundings

We are given the enthalpy change 螖H鈦� as +20.0 kJ. Since the value is positive, it means that the reaction is endothermic. Therefore, the reaction absorbs heat from its surroundings, leading to a decrease in the randomness or disorder of the surroundings.
03

Calculating Gibbs free energy change (螖G鈦�) at 298 K

We can use the relation between Gibbs free energy change, enthalpy change, and entropy change (螖G鈦� = 螖H鈦� - T螖S鈦�) to calculate 螖G鈦� at 298 K. Given : 螖H鈦� = +20.0 kJ 螖S鈦� = +100.0 J/K T = 298 K First, we need to convert 螖H鈦� and 螖S鈦� to the same units. Here, we will convert 螖H鈦� to J by multiplying it by 1000: 螖H鈦� = 20.0 kJ 脳 1000 J/kJ = 20,000 J Now we can use the formula: 螖G鈦� = 螖H鈦� - T螖S鈦� 螖G鈦� = 20,000 J - (298 K 脳 100 J/K)
04

Calculate 螖G鈦�

By substituting the values, we get: 螖G鈦� = 20,000 J - 29,800 J 螖G鈦� = -9,800 J
05

Determine if the reaction is spontaneous at 298 K

Since the Gibbs free energy change 螖G鈦� is negative (-9,800 J), the reaction is spontaneous at 298 K under standard conditions.

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Key Concepts

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

Entropy Change
Entropy change refers to the variation in the randomness or disorder of a system during a chemical reaction. In thermodynamics, it is represented as \( \Delta S \). When \( \Delta S \) is positive, it indicates an increase in disorder. Conversely, a negative \( \Delta S \) implies a decrease in disorder.
In this exercise, the provided \( \Delta S^{\circ} = +100.0 \ \mathrm{J/K} \) suggests a positive entropy change. This tells us that the products of the reaction are more random than the reactants. As a result, the system's entropy increases, making the molecules more disordered.
This information can be vital when predicting the behavior of chemical reactions.
  • A positive \( \Delta S \) could mean gases are being formed, increasing randomness compared to liquids or solids.
  • It helps in assessing spontaneity together with enthalpy and temperature.
Enthalpy Change
Enthalpy change involves the heat change of a system during a reaction and is denoted as \( \Delta H \). It signifies whether heat is absorbed or released.
A positive \( \Delta H \) indicates that the reaction is endothermic, absorbing heat from the surroundings. A negative \( \Delta H \) means it is exothermic, releasing heat. In this exercise, the enthalpy change \( \Delta H^{\circ} = +20.0 \ \mathrm{kJ} \) suggests the reaction absorbs energy.
This absorption can cause the surroundings to cool down due to reduced thermal motion.
  • This reaction's positive \( \Delta H \) leads to a decrease in the surroundings' disorder, as energy absorption often decreases molecular motion.
  • Understanding \( \Delta H \) helps in determining reaction feasibility and conditions required for the reaction to proceed.
Spontaneity of Reaction
The spontaneity of a reaction determines if it can proceed without external influence. It's assessed using Gibbs free energy change \( \Delta G \).
The expression \( \Delta G = \Delta H - T \Delta S \) helps us evaluate this. If \( \Delta G \) is negative, the reaction is spontaneous; if positive, it's non-spontaneous.
For this reaction, calculating \( \Delta G^{\circ} \) at \( 298 \ \mathrm{K} \): \[ \Delta G^{\circ} = +20,000 \ \mathrm{J} - (298 \ \mathrm{K} \times 100 \ \mathrm{J/K}) = -9,800 \ \mathrm{J} \]
The result indicates that the reaction is spontaneous under standard conditions. This is because:
  • A negative \( \Delta G \) suggests the process is thermodynamically favorable.
  • The reaction can progress without needing additional energy inputs.
Understanding spontaneity is crucial for determining if a reaction can occur naturally and how conditions like temperature influence it.

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Most popular questions from this chapter

Today, most candles are made of paraffin wax. A typical component of paraffin wax is the hydrocarbon \(\mathrm{C}_{31} \mathrm{H}_{64}\) which is solid at room temperature. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{31} \mathrm{H}_{64}(s)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) ( \(\mathbf{b}\) ) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

Using the data in Appendix \(C\) and given the pressures listed, calculate \(K_{\mathrm{p}}\) and \(\Delta G\) for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \\ \quad R_{\mathrm{N}_{2}}=263.4 \mathrm{kPa}, P_{\mathrm{H}_{2}}=597.8 \mathrm{kPa}, P_{\mathrm{NH}_{3}}=101.3 \mathrm{kPa} \\ \text { (b) } 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ \(P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.07 \mathrm{kPa}\) $$ \begin{array}{l} \quad R_{\mathrm{N}_{2}}=50.7 \mathrm{kPa}, P_{\mathrm{H}_{2} \mathrm{O}}=30.4 \mathrm{kPa} \\ \text { (c) } \mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \\ P_{\mathrm{N}_{2} \mathrm{H}_{4}}=101.3 \mathrm{kPa}, P_{\mathrm{N}_{2}}=152.0 \mathrm{kPa}, P_{\mathrm{H}_{2}}=253.3 \mathrm{kPa} \end{array} $$

The normal boiling point of \(n\) -octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is \(125^{\circ} \mathrm{C}\). (a) Is the condensation of gaseous \(n\) -octane to liquid \(n\) -octane an endothermic or exothermic process? (b) In what temperature range is the boiling of \(n\) -octane a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid \(n\) -octane and gaseous \(n\) -octane are in equilibrium? Explain.

Indicate whether each statement is true or false. (a) Unlike enthalpy, where we can only ever know changes in \(H,\) we can know absolute values of \(S .(\mathbf{b})\) If you heat a gas such as \(\mathrm{CO}_{2}\), you will increase its degrees of translational, rotational and vibrational motions. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass. At a given temperature, they will have the same number of microstates.

An ice cube with a mass of \(25 \mathrm{~g}\) at \(-18{ }^{\circ} \mathrm{C}\) (typical freezer temperature) is dropped into a cup that holds \(250 \mathrm{~mL}\) of hot water, initially at \(85^{\circ} \mathrm{C}\). What is the final temperature in the cup? The density of liquid water is \(1.00 \mathrm{~g} / \mathrm{mL}\); the specific heat capacity of ice is \(2.03 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C} ;\) the specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}-\mathrm{K} ;\) the enthalpy of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol}\).
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