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For the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\) all but one of the following equations is correct. Which is incorrect, and why? (a) \(K=K_{\mathrm{p}} ;\) (b) \(\Delta S^{\circ}=\) \(\left(\Delta G^{\circ}-\Delta H^{\circ}\right) / T ;\left(\text { c) } K_{\mathrm{p}}=e^{-\Delta G^{\circ} / R T} ;(\mathrm{d}) \Delta G=\Delta G^{\circ}+\right.\) \(R T \ln Q\).

Short Answer

Expert verified
The incorrect equation is (a) \(K=K_{\mathrm{p}}\). The equilibrium constant \(K\) is not equal to the equilibrium constant in terms of pressure (\(K_{p}\)) for this reaction, because the difference in moles of gaseous reactants and products is not zero.

Step by step solution

01

Analyze equation (a)

The equation is \(K=K_{p}\). This equation is incorrect. The equilibrium constant \(K\) is equal to the equilibrium constant in terms of pressure (\(K_{p}\)) only when Δn, the difference in moles of gaseous reactants and products, is equal to zero. For the given reaction 2NO(g) + O2(g) -> 2NO2(g), Δn is not zero (4-2 = 2), so \(K\) is not equal to \(K_{p}\). Therefore, the equation (a) is incorrect.
02

Analyze equation (b)

The equation is \(\Delta S^{\circ}=\left(\Delta G^{\circ}-\Delta H^{\circ}\right) / T\). This is the correct way to express standard entropy using Gibbs’ equation. The standard entropy change (\(\Delta S^{0}\)) is indeed equal to the standard free energy change (\(\Delta G^{0}\)) minus the standard enthalpy change (\(\Delta H^{0}\)), all divided by the absolute temperature (T).
03

Analyze equation (c)

The equation is \(K_{p}=e^{-\Delta G^{\circ} / R T}\). This equation is correct; it comes from manipulating the equation for the standard Gibbs free energy change, which is -RT ln(K). Kp is indeed equal to \(e^{-\Delta G^{\circ} / R T}\).
04

Analyze equation (d)

The equation is \(\Delta G = \Delta G^{\circ} + R T \ln Q\). This is correct. The Gibbs free energy can indeed be written this way, as it takes into account the standard free energy change (\(\Delta G^{\circ}\)), the universal gas constant (R), temperature (T) and the Reaction Quotient (Q).

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, represented by the symbol \( K \), is a numerical value that expresses the relative proportions of reactants to products in a chemical reaction at equilibrium. It is a crucial concept in the study of chemical reactions and provides insight into the degree to which a reaction proceeds.
For any given reaction at a constant temperature, the equilibrium constant remains the same irrespective of the initial concentrations of reactants and products. This constant can be expressed in terms of concentrations, denoted as \( K_c \), or in terms of partial pressures, denoted as \( K_p \). These two are related but are not always equal as they depend on the change in the number of moles of gas, represented by \( \Delta n \), when going from reactants to products. If \( \Delta n \) equals zero, then \( K \) would be equal to \( K_p \), otherwise, they must be related through an equation involving \( \Delta n \) and the universal gas constant \( R \).
When a student approaches a problem involving equilibrium constants, they must pay close attention to the reaction involved and consider whether they are dealing with concentrations or pressures. It's essential to use the correct form of the equilibrium constant to avoid inaccuracies.
Gibbs Free Energy
Gibbs free energy, symbolized as \( G \), is a thermodynamic quantity that serves as a useful indicator of whether a process, including chemical reactions, will proceed spontaneously. A negative value for the change in Gibbs free energy, \( \Delta G \), indicates that the reaction will proceed spontaneously in the forward direction under constant pressure and temperature.
Moreover, the relationship between Gibbs free energy and equilibrium can be expressed mathematically. At equilibrium, \( \Delta G = 0 \), and this ties into the other key formulas. For example, \( \Delta G^\circ = -RT \ln(K) \) connects the standard Gibbs free energy change to the equilibrium constant \( K \). The Gibbs free energy change for a reaction that is not at equilibrium can be calculated using this relationship: \( \Delta G = \Delta G^\circ + RT \ln(Q) \), where \( Q \) is the reaction quotient.
Understanding Gibbs free energy is fundamental in predicting the direction and extent of chemical reactions. In the context of equilibrium, it can offer insights into the effects of temperature changes on the position of equilibrium.
Reaction Quotient
The reaction quotient, \( Q \), represents the ratio of the concentrations (or partial pressures) of products to reactants at any point in time for a reaction. It serves as a predictor to determine which direction a reaction will shift to reach equilibrium. Its expression is identical to that of the equilibrium constant \( K \), but while \( K \) is exclusively for systems at equilibrium, \( Q \) can be applied to any moment of the reaction process.
If \( Q < K \), the reaction will proceed forward to produce more products and reach equilibrium. Conversely, if \( Q > K \), there will be a shift towards the formation of more reactants. When \( Q = K \), the reaction is at equilibrium, and no further shift will occur. The concept of the reaction quotient is central when using the equation \( \Delta G = \Delta G^\circ + RT \ln(Q) \) because it informs about the spontaneity and directionality of reactions not at equilibrium.
Entropy
Entropy, symbolized as \( S \), is a measure of the disorder or randomness within a thermodynamic system. In the realm of chemical reactions, it is often associated with the distribution of energy and particles. An increase in entropy typically corresponds to a more disordered or less structured system.
The change in entropy, \( \Delta S \), plays a vital role in determining the spontaneity of a reaction alongside enthalpy and temperature. The relationship between Gibbs free energy, entropy, and enthalpy is given by the equation \( \Delta G = \Delta H - T\Delta S \), where \( \Delta H \) is the change in enthalpy, \( T \) is the temperature, and \( \Delta G \) is the change in Gibbs free energy.
An understanding of entropy allows students to grasp the underlying reasons behind energy distribution and the directional flow of chemical processes, as well as the balance between enthalpic and entropic contributions to the spontaneity of reactions.

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Most popular questions from this chapter

The reaction, \(2 \mathrm{Cl}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{Cl}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \Delta H=\) \(-161 \mathrm{kJ},\) is expected to be (a) spontaneous at all temperatures; (b) spontaneous at low temperatures, but nonspontaneous at high temperatures; (c) nonspontaneous at all temperatures; (d) spontaneous at high temperatures only.

Arrange the entropy changes of the following processes, all at \(25^{\circ} \mathrm{C},\) in the expected order of increasing \(\Delta S,\) and explain your reasoning: (a) \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{atm}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 1 \mathrm{atm})\) (b) \(\mathrm{CO}_{2}(\mathrm{s}, 1 \mathrm{atm}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}, 10 \mathrm{mm} \mathrm{Hg})\) (c) \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{atm}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 10 \mathrm{mmHg})\)

Use ideas from this chapter to explain this famous remark attributed to Rudolf Clausius (1865)\(:^{\prime \prime} \mathrm{Die}\) Energie der Welt ist konstant; die Entropie der Welt strebt einem Maximum zu." ("The energy of the world is constant; the entropy of the world increases toward a maximum.")

Consider the vaporization of water: \(\mathrm{H}_{2} \mathrm{O}(1) \longrightarrow\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100^{\circ} \mathrm{C},\) with \(\mathrm{H}_{2} \mathrm{O}(1)\) in its standard state, but with the partial pressure of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(2.0 \mathrm{atm}\) Which of the following statements about this vaporization at \(100^{\circ} \mathrm{C}\) are true? (a) \(\Delta G^{\circ}=0,\) (b) \(\Delta G=0\) (c) \(\Delta G^{\circ}>0,\) (d) \(\Delta G>0 ?\) Explain.

To establish the law of conservation of mass, Lavoisier carefully studied the decomposition of mercury(II) oxide: $$\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=+90.83 \mathrm{kJ}\) and \(\Delta G^{\circ}=+58.54 \mathrm{kJ}\) (a) Show that the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in equilibrium with \(\mathrm{HgO}(\mathrm{s})\) and \(\mathrm{Hg}(\mathrm{l})\) at \(25^{\circ} \mathrm{C}\) is extremely low. (b) What conditions do you suppose Lavoisier used to obtain significant quantities of oxygen?

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