Calculate \(\Delta S_{\text {sur }}\) for the following reactions at \(25^{\circ} \mathrm{C}\) and 1 atm. $$\text { a. } \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$$$$\begin{aligned} &\Delta H^{\circ}=-2221 \mathrm{kJ}\\\ &\text { b. } 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \quad \Delta H^{\circ}=112 \mathrm{kJ} \end{aligned}$$

Short Answer

Expert verified
The change in entropy of the surroundings for reaction (a) is 7453 J/mol·K and for reaction (b) is -376 J/mol·K.

Step by step solution

01

Convert the given temperature to kelvin

To convert 25°C to kelvin, add 273.15. \[25^{\circ}\text{C} + 273.15 = 298.15\, \text{K}\] Temperature, T = 298.15 K
02

Calculate \(\Delta S_{\text{sur}}\) for reaction (a)

Using the given \(\Delta H^{\circ} = -2221\, \text{kJ}\) for reaction (a) and temperature T = 298.15 K, we can plug these values into the formula for \(\Delta S_{\text{sur}}\): \[\Delta S_{\text{sur}} = -\frac{-2221\, \text{kJ}}{298.15\, \text{K}}\] Convert \(\Delta H^{\circ}\) to joules: \(-2221\, \text{kJ} = -2221000\, \text{J}\) \[\Delta S_{\text{sur}} = -\frac{-2221000\, \text{J}}{298.15\, \text{K}}\]
03

Calculate \(\Delta S_{\text{sur}}\) for reaction (b)

Using the given \(\Delta H^{\circ} = 112\, \text{kJ}\) for reaction (b) and temperature T = 298.15 K, we can plug these values into the formula for \(\Delta S_{\text{sur}}\): \[\Delta S_{\text{sur}} = -\frac{112\, \text{kJ}}{298.15\, \text{K}}\] Convert \(\Delta H^{\circ}\) to joules: \(112\, \text{kJ} = 112000\, \text{J}\) \[\Delta S_{\text{sur}} = -\frac{112000\, \text{J}}{298.15\, \text{K}}\]
04

Express the results in J/mol·K

Now, we can calculate \(\Delta S_{\text{sur}}\) for each reaction: For reaction (a): \[\Delta S_{\text{sur}} = \frac{2221000\, \text{J}}{298.15\, \text{K}} = 7453\, \text{J/mol·K}\] For reaction (b): \[\Delta S_{\text{sur}} = -\frac{112000\, \text{J}}{298.15\, \text{K}} = -376\, \text{J/mol·K}\] Therefore, \(\Delta S_{\text{sur}}\) for reaction (a) is 7453 J/mol·K and for reaction (b) is -376 J/mol·K.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics dealing with heat, work, temperature, and the related properties of matter. The field explores how different forms of energy can be transformed into one another and how they affect matter. It's fundamental in explaining why physical and chemical phenomenon occur in a certain way.

At the heart of thermodynamics are four laws which can be applied to all sorts of thermodynamic systems, albeit closed or open. One of the key principles related to the given exercise is the second law of thermodynamics, which states that the entropy of an isolated system always increases or remains constant. In simple terms, entropy is a measure of the disorder or randomness in a system, and this concept helps us understand the irreversible nature of natural processes. When we calculate the change in entropy (abla S) during a chemical reaction, we are assessing how the disorder of the system and its surroundings has been altered by that reaction.
Chemical Reactions
Chemical reactions involve the transformation of one set of chemical substances to another through the breaking and forming of chemical bonds. These reactions are influenced by various thermodynamic properties, such as the internal energy, enthalpy (Delta H), entropy (Delta S), and temperature (T), all of which determine the spontaneity and extent of the reaction.

Different types of chemical reactions, such as combustion in reaction (a) from the exercise or the decomposition in reaction (b), involve changes in enthalpy and entropy. Exothermic reactions, like combustion, release heat and typically have a negative Delta H value, which reflects the release of energy. In contrast, endothermic reactions, such as the decomposition of NO2, absorb heat, resulting in a positive Delta H. These changes, combined with the temperature of the surroundings, allow us to calculate the change in the entropy of the surroundings, further revealing insights into the second law of thermodynamics.
Gibbs Free Energy
Gibbs free energy (G) integrates enthalpy, entropy, and temperature to predict the direction of a chemical reaction and whether it’s spontaneous. The Gibbs free energy change (Delta G) for a process is defined as: Delta G = Delta H - TDelta S, where Delta H is the change in enthalpy, T is the absolute temperature in Kelvin (K), and Delta S is the change in entropy of the system. A spontaneous reaction occurs when Delta G is negative, indicating that the process can proceed without the input of additional energy.

When we apply this to chemical reactions, a negative Delta G suggests a reaction is likely to occur, while a positive value suggests it is not favored under the given conditions. For example, the calculation in the exercise helps us understand the entropy change of the surroundings (Delta S_sur), which is directly related to the concept of Gibbs free energy. By measuring these changes, scientists can better predict the behavior of chemical reactions in various conditions.

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Most popular questions from this chapter

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a \(100 \%\) yield of pure \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

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.

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

Consider the reaction $$2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S .\) b. Would the reaction be more spontaneous at high or low temperatures?

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 \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

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