(a) Calculate the reaction free energy of \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) when the partial pressures of \(\mathrm{N}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\) are \(1.0,4.2\), and 63 bar, respectively, and the temperarure is \(400 \mathrm{~K}\). For this reaction, \(K=41\) at \(400 \mathrm{~K}\). (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

Short Answer

Expert verified
The reaction quotient Q is found to be different from K, suggesting the reaction is not at equilibrium. Depending on the calculated value of Q compared to K, this will indicate the direction in which the reaction will proceed.

Step by step solution

01

- Use the Reaction Quotient

Calculate the reaction quotient, Q, using the given partial pressures. The reaction quotient Q is defined as the product of the partial pressures of the products raised to their stoichiometric coefficients divided by the product of the partial pressures of the reactants raised to their stoichiometric coefficients. For the balanced reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\), Q is given by \(Q = \frac{P_{\mathrm{NH}_3}^2}{P_{\mathrm{N}_2} (P_{\mathrm{H}_2})^3}\).
02

- Calculate the Reaction Quotient, Q

Plug the partial pressures into the equation for Q: \(Q = \frac{(63)^2}{(1.0)(4.2)^3}\). Calculate Q to compare with the equilibrium constant K.
03

- Calculate the Reaction Free Energy Change

The reaction free energy change, \(\Delta G\), can be calculated using the equation \(\Delta G = \Delta G^\circ + RT \ln (Q)\), where \(\Delta G^\circ\) is the standard reaction free energy change, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient. Since \(K = e^{-\Delta G^\circ / (RT)}\), one can find \(\Delta G^\circ\) using \(\Delta G^\circ = -RT \ln (K)\).
04

- Calculate Standard Reaction Free Energy Change

First, calculate \(\Delta G^\circ\) using the equilibrium constant K and the given temperature: \(\Delta G^\circ = -RT \ln (K)\). The value of the gas constant R is \(8.314 J/(mol\cdot K)\).
05

- Calculate the Actual Reaction Free Energy Change

After finding \(\Delta G^\circ\), use it to calculate \(\Delta G\): \(\Delta G = \Delta G^\circ + RT \ln (Q)\). This will give the reaction free energy at the given conditions.
06

- Analyze Q relative to K

Compare the calculated Q to the equilibrium constant K to determine if the reaction will proceed towards products, towards reactants, or is at equilibrium: If \(Q < K\), the reaction will form more products; if \(Q > K\), it will form more reactants; if \(Q = K\), the reaction is at equilibrium.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. As a result, there's no net change in the concentration of reactants and products over time. Equilibrium does not mean that reactants and products are present in equal amounts – rather, it indicates that their ratios are constant.

When a system reaches equilibrium, it may appear to be static, but in reality, both reactants and products are continually being formed and consumed at equal rates. Understanding equilibrium is crucial in predicting the behavior of chemical reactions and formulating products in various industries.
Reaction Quotient (Q)
The reaction quotient, Q, is a mathematical expression used to determine the direction in which a reaction will proceed at any point in time. It takes into account the current concentrations or partial pressures of the reactants and products. The value of Q can be compared with the equilibrium constant, K, to predict the shift that will occur in order to reach equilibrium.

For gaseous reactions, the reaction quotient is calculated using partial pressures. It is given by the partial pressures of the products raised to the power of their stoichiometric coefficients, divided by the partial pressures of the reactants raised to their respective powers. If Q is less than K, the reaction will proceed in the forward direction to form more products. If Q is greater than K, the reaction will shift to form more reactants.
Equilibrium Constant (K)
The equilibrium constant, K, is a number that expresses the ratio of the concentration of products to reactants for a reversible reaction at chemical equilibrium and a constant temperature. K is constant for a given reaction at a specific temperature and is independent of the initial concentrations of reactants and products.

In the application, if the reaction quotient Q equals the equilibrium constant K, the reaction is at equilibrium. The value of K can help predict whether a reaction mixture will proceed to form more products or reactants as it attempts to reach equilibrium. K is determined experimentally and varies with temperature, so it’s essential to ensure that K and the reaction conditions (such as temperature) correspond accordingly.
Gibbs Free Energy
Gibbs free energy, denoted as \(\Delta G\), is a thermodynamic quantity that predicts the direction of a chemical reaction. For reactions occurring at constant temperature and pressure, a negative \(\Delta G\) indicates that a process is spontaneous, meaning it can occur without any added energy. Conversely, a positive \(\Delta G\) implies that the process is non-spontaneous.

\(\Delta G\) also relates closely to the concepts of equilibrium and reaction quotient. It can be calculated from the standard Gibbs free energy change (\(\Delta G^\circ\)), which is based on standard conditions, along with the actual concentration or pressure of reactants and products, encapsulated in the reaction quotient (Q). The relationship between \(\Delta G\), \(\Delta G^\circ\), and Q is used to determine the spontaneity of the reaction under non-standard conditions.
Partial Pressures
Partial pressures are used in chemistry to describe the pressure exerted by a single gas when it is part of a mixture of gases. Each gas in a mixture exerts pressure as if it were the only gas present, and the total pressure of the mixture is the sum of the partial pressures of each individual gas.

Partial pressures are particularly relevant in the context of gaseous equilibria, where they are used to calculate the reaction quotient (Q) and equilibrium constants (K). When dealing with reactions involving gases, changes in partial pressures can affect both Q and K, thus influencing the direction in which the equilibrium will shift. Furthermore, the concept of partial pressures is also essential in applications ranging from respiratory physiology to the design of industrial chemical processes.

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Most popular questions from this chapter

Calculate the standard reaction free energy of each of the following reactions: (a) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), K=41\) ar \(400 \mathrm{~K}\) (b) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.0 \times 10^{4}\) at \(700 \mathrm{~K}\)

For the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\), \(K_{\mathrm{c}}=160\) at \(500 \mathrm{~K}\). An analysis of a reaction mixture at \(500 \mathrm{~K}\) showed that it had the composition \(4.8 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}, 2.4 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{I}_{2}\), and \(2.4 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{HI}\). (a) Calculate the reaction quoticnt. (b) Is the reaction mixture at equilibrium? (c) If not, is there a tendency to form more reactants or more products?

A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

The photosynthesis reaction is \(6 \mathrm{CO}_{2}(\mathrm{~g})+\) \(6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_{\mathrm{g}} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{~g})\), and \(\Delta H^{\circ}=+2802 \mathrm{~kJ}\). Suppose that the reaction is at cquilibrium. State the effect (tend to shift toward the formation of reactants, tend to shift toward the formation of products, or have no effect) that each of the following changes will have on the equilibrium composition: (a) the partial pressure of \(\mathrm{O}_{2}\) is increased; (b) the system is compressed; (c) the amount of \(\mathrm{CO}_{2}\) is increased; (d) the temperature is increased; (c) some of the \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is removed; \((f)\) water is added; \((g)\) the partial pressure of \(\mathrm{CO}_{2}\) is decreased.

Predict whether each of the following equilibria will shift toward products or reactants with a temperature increase. (a) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)=2 \mathrm{NO}_{2}(g), \Delta H^{\circ}=+57 \mathrm{~kJ}\) (b) \(\mathrm{X}_{2}(\mathrm{~g})=2 \mathrm{X}(\mathrm{g})\), where \(\mathrm{X}\) is a halogen (c) \(\mathrm{Ni}(\mathrm{s})+4 \mathrm{CO}(\mathrm{g})=\mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{~g}), \Delta \mathrm{H}^{+}=-161 \mathrm{~kJ}\) (d) \(\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), \(\Delta H^{\circ}=-90 \mathrm{~kJ}\)

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