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Chapter 17: Problem 64

Consider the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g) \quad K=4.4 \times 10^{-19} $$ (a) Calculate \(\Delta G^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\). (b) Calculate \(\Delta G_{i}^{\circ}\) for \(\mathrm{N}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Question: Calculate (a) the standard Gibbs free energy change (螖G掳) for the given reaction and (b) the standard Gibbs free energy of formation (螖G岬⒙�) for N2O at 25掳C, given the following information: Reaction: N2O + NO2 鈫� 3NO K = 4.4 脳 10鈦宦光伖 螖G掳(NO) = 87.5 kJ/mol 螖G掳(NO2) = 52.1 kJ/mol Answer: (a) 螖G掳 for the reaction at 25掳C = 2.67 脳 10虏 kJ/mol (b) 螖G岬⒙� for N2O at 25掳C = 8.4 kJ/mol

Step by step solution

01

Part (a) - Calculate \(\Delta G^{\circ}\) for the reaction at 25掳C

To calculate \(\Delta G^{\circ}\) for the reaction, we will use the formula connecting the equilibrium constant (K) with the standard Gibbs free energy change: $$ \Delta G^{\circ} = -RT \ln K $$ where \(R\) is the gas constant (8.314 J/mol K), \(T\) is the temperature in Kelvin (298 K) and \(K\) is the equilibrium constant given as \(4.4 \times 10^{-19}\). Plug in the values and solve for \(\Delta G^{\circ}\): $$ \Delta G^{\circ} = - (8.314 \,\text{J/mol K})(298 \,\text{K}) \ln (4.4 \times 10^{-19}) $$ $$ \Delta G^{\circ} \approx 2.67 \times 10^{2} \,\text{kJ/mol} $$
02

Part (b) - Calculate \(\Delta G_{i}^{\circ}\) for N2O at 25掳C

To calculate the standard Gibbs free energy of formation for N2O, we will use the equation for the standard Gibbs free energy change in terms of the standard Gibbs free energies of formation of the products and reactants: $$ \Delta G^{\circ} = \sum_\text{products}(\text{nG}) - \sum_\text{reactants}(\text{nG}) $$ For the given reaction: $$ \Delta G^{\circ} = 3 \Delta G_{\mathrm{NO}}^{\circ} - (\Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} + \Delta G_{\mathrm{NO}_{2}}^{\circ}) $$ We are given the standard Gibbs free energy of formation for NO2 and NO: $$ \Delta G_{\mathrm{NO}_{2}}^{\circ} = 52.1 \,\text{kJ/mol} $$ $$ \Delta G_{\mathrm{NO}}^{\circ} = 87.5 \,\text{kJ/mol} $$ Insert the values and the calculated \(\Delta G^{\circ}\) from part (a): $$ 2.67 \times 10^{2} \,\text{kJ/mol} = 3 (87.5 \,\text{kJ/mol}) - (\Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} + 52.1 \,\text{kJ/mol}) $$ Now, solve for \(\Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ}\): $$ \Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} = 3 (87.5 \,\text{kJ/mol}) - 2.67 \times 10^{2} \,\text{kJ/mol} - 52.1 \,\text{kJ/mol} $$ $$ \Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} \approx 8.4 \,\text{kJ/mol} $$

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Key Concepts

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

Understanding Equilibrium Constant
The equilibrium constant, often represented as \( K \), is a fundamental concept in chemical reactions. It describes the ratio of the concentrations of products to reactants at equilibrium. When a reaction reaches equilibrium, the forward and reverse reaction rates are equal. This balance means the concentrations of the reactants and products remain constant over time, but it's essential to note that they do not have to be equal to each other.

In the expression for equilibrium constant:
  • For a reaction \( aA + bB \leftrightarrow cC + dD \), the equilibrium constant \( K \) is given by \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \).
  • The brackets \([ ]\) denote concentration in terms of molarity (moles per liter).
This ratio is temperature dependent. As seen with your given reaction, even though the \( K \) value is extremely small (\( 4.4 \times 10^{-19} \)), it tells us that at equilibrium, the concentration of products (NO) is very low compared to reactants (N鈧侽 and NO鈧�). Such a tiny number indicates that the reaction does not spontaneously form many products under the given conditions.
Thermodynamics and Gibbs Free Energy
Thermodynamics is the study of energy transformations. Gibbs free energy, denoted by \( \Delta G \), is a crucial thermodynamic quantity that indicates the spontaneity of a chemical reaction. The equation \( \Delta G^{\circ} = -RT \ln K \) directly relates Gibbs free energy to the equilibrium constant \( K \). This equation uses the gas constant \( R \) (8.314 J/mol K) and the absolute temperature \( T \).

The relationship can be summarized as follows:
  • If \( \Delta G^{\circ} < 0 \), the reaction is spontaneous and will proceed in the forward direction.
  • If \( \Delta G^{\circ} = 0 \), the system is at equilibrium.
  • If \( \Delta G^{\circ} > 0 \), the reaction is non-spontaneous and tends to go in the reverse direction.
For the given reaction at 25掳C (which is \( 298 \) Kelvin), the positive \( \Delta G^{\circ} \approx 2.67 \times 10^{2} \,\text{kJ/mol} \) indicates that at standard conditions, the reaction is not spontaneous. This aligns with the very low equilibrium constant, showing that energy is needed for the formation of significant amounts of product.
The Dynamics of Chemical Reactions
Chemical reactions involve the transformation of reactants into products. Understanding a reaction's dynamics, such as rate and extent, depends heavily on concepts like reaction mechanisms, energy changes, and ultimately the equilibrium constant.

Our exercise involves a seemingly simple reaction: \( \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \rightarrow 3 \mathrm{NO}(g) \). Though it looks straightforward, reactions often occur in complex steps that influence how and when equilibrium is reached.
  • The rates at which reactants are converted to products depend on factors like temperature, pressure, and catalysts.
  • The energy changes associated with reactions are explained using Gibbs free energy, which considers both enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)).
  • The ability to predict reaction spontaneity and direction heavily relies on thermodynamics, as well as the calculated \( \Delta G \).
In summary, while chemical reactions are driven towards equilibrium, understanding them fully requires an interplay of kinetics, thermodynamics, and molecular behavior. This deeper insight is what guides scientists to optimize conditions in various industrial and laboratory processes.

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Most popular questions from this chapter

At \(1200 \mathrm{~K}\), an equilibrium mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) gases contains 98.31 mol percent CO and some solid carbon. The total pressure of the mixture is \(1.00 \mathrm{~atm}\). For the system $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) $$ calculate (a) \(P_{\mathrm{Co}}\) and \(P_{\mathrm{CO}_{2}}\) (b) \(K\) (c) \(\Delta G^{\circ}\) at \(1200 \mathrm{~K}\)

A student is asked to prepare a \(0.030 \mathrm{M}\) aqueous solution of \(\mathrm{PbCl}_{2}\) (a) Is this possible at \(25^{\circ} \mathrm{C} ?\) (Hint: Is dissolving \(0.030 \mathrm{~mol}\) of \(\mathrm{PbCl}_{2}\) at \(25^{\circ} \mathrm{C}\) possible? \()\) (b) If the student used water at \(100^{\circ} \mathrm{C}\), would this be possible?

Pencil "lead" is almost pure graphite. Graphite is the stable elemental form of carbon at \(25^{\circ} \mathrm{C}\) and 1 atm. Diamond is an allotrope of graphite. Given $$ \text { diamond: } \Delta H_{\mathrm{f}}^{\circ}=1.9 \mathrm{~kJ} / \mathrm{mol} ; S^{\circ}=2.4 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} $$ at what temperature are the two forms in equilibrium at 1 atm? \(\mathrm{C}\) (graphite) \(\rightleftharpoons \mathrm{C}\) (diamond)

Given the following data for sodium $$ \begin{aligned} &\mathrm{Na}(s): S^{\circ}=51.2 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\\ &\mathrm{Na}(g): S^{\circ}=153.6 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \quad \Delta H_{\mathrm{f}}^{\circ}=108.7 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ estimate the temperature at which sodium sublimes at 1 atm. $$ \mathrm{Na}(s) \rightleftharpoons \mathrm{Na}(g) $$

Red phosphorus is formed by heating white phosphorus. Calculate the temperature at which the two forms are at equilibrium, given $$ \text { white } \text { P: } \Delta H_{\mathrm{f}}^{\circ}=0.00 \mathrm{~kJ} / \mathrm{mol} ; S^{\circ}=41.09 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} $$ red \(\mathrm{P}: \Delta H_{\mathrm{f}}^{\circ}=-17.6 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{S}^{\circ}=22.80 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\).
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