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Consider the following reaction: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of \(\pm 1\) in all quantities): a. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{~atm}\) b. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=200 \mathrm{~atm}, P_{\mathrm{H}_{2}}=600 \mathrm{~atm}\), \(P_{\mathrm{NH}_{3}}=200 \mathrm{~atm}\)

Short Answer

Expert verified
For case a, the reaction quotient, \(Q_a = \frac{(50)^2}{200 \times (200)^3}\). Using the equation \(\Delta G = \Delta G^{\circ} + RT \ln Q\), we can calculate \(\Delta G_a\). Similarly, for case b, the reaction quotient, \(Q_b = \frac{(200)^2}{200 \times (600)^3}\), and we can calculate \(\Delta G_b\). After finding the values for \(\Delta G_a\) and \(\Delta G_b\), report the results with their uncertainties of ±1.

Step by step solution

01

Calculate Q for each case#a.#

Given the pressures: \(P_{\mathrm{N}_{2}}=200\: \mathrm{atm}\), \(P_{\mathrm{H}_{2}}=200\: \mathrm{atm}\), \(P_{\mathrm{NH}_{3}}=50\: \mathrm{atm}\) The reaction quotient, Q, is given by the formula: \[ Q = \frac{P_{\mathrm{NH}_{3}}^2}{P_{\mathrm{N}_{2}} \times P_{\mathrm{H}_{2}}^3} \] By substituting the given pressures, we have: \[ Q_a = \frac{(50)^2}{200 \times (200)^3} \]
02

b.#

Given the pressures: \(P_{\mathrm{N}_{2}}=200\: \mathrm{atm}\), \(P_{\mathrm{H}_{2}}=600\: \mathrm{atm}\), \(P_{\mathrm{NH}_{3}}=200\: \mathrm{atm}\) Using the Q formula: \[ Q_b = \frac{(200)^2}{200 \times (600)^3} \]
03

Find ΔG°#.#

You are not given the value of ΔG° in this problem. However, it can be found in a textbook or reference resource, or you may have been provided with this information elsewhere in your course materials. For this exercise, we will assume a value of ΔG° = -33 kJ/mol. Note: If the problem assumes that the reaction occurs at the standard state, we must first compute the standard state equilibrium constant, K, using the ΔG° value, which will be used later for the calculations: \[ K = \exp \left({-\frac{\Delta G^{\circ}}{RT}}\right) \]
04

Calculate ΔG for each case#a.#

Given \(T = 298 \mathrm{~K}\), \(R = 8.314 \mathrm{~J/(mol\: K)}\), and the calculated value of \(Q_a\). Now, we can use the equation: \[ \Delta G = \Delta G^{\circ} + RT \ln Q_a \] \[ \Delta G_a = -33000\: \mathrm{J/mol} + (8.314\: \mathrm{J/(mol \: K)}) (298\: \mathrm{K}) \ln Q_a \]
05

b.#

Using the same T, R values and the calculated value of \(Q_b\), we have: \[ \Delta G = \Delta G^{\circ} + RT \ln Q_b \] \[ \Delta G_b = -33000\: \mathrm{J/mol} + (8.314\: \mathrm{J/(mol \: K)}) (298\: \mathrm{K}) \ln Q_b \] Now, you can plug in the values of Q from Steps 1a and 1b to find the ΔG values for cases a and b, and report the results with their uncertainties of ±1.

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Key Concepts

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

Reaction Quotient (Q)
The concept of the reaction quotient, Q, is pivotal in understanding the direction in which a chemical reaction will proceed at any given moment. In essence, Q compares the concentrations or partial pressures of products and reactants at a non-equilibrium state. It is represented by the formula:

\[\begin{equation}Q = \frac{[\text{products}]}{[\text{reactants}]}\end{equation}\]

with the concentrations raised to the power of their coefficients in the balanced chemical equation. When Q is calculated under different conditions, as in the exercise, it helps predict if the reaction will shift towards products or reactants to reach equilibrium.
  • For Q
  • When Q>K, the reaction will reverse, forming more reactants.
  • If Q=K, the system is at equilibrium, and no net change will occur.
Having the Q value is a stepping stone towards calculating other variables of interest in chemical reactions, such as the Gibbs free energy change (ΔG).
Gibbs Free Energy (\(\Delta G\))
Gibbs free energy (ΔG) is a thermodynamic quantity that predicts the direction of chemical reactions and determines whether a process is spontaneous. A negative ΔG indicates a spontaneous reaction, positive ΔG means a reaction is non-spontaneous, and a ΔG of zero denotes a system at equilibrium.

Using the equation:\[\begin{equation}\Delta G = \Delta G^\circ + RT\ln Q\end{equation}\]

it is possible to calculate the change in Gibbs free energy under any set of conditions if the standard free energy change (ΔG°) and the reaction quotient (Q) are known. In the standard equation, T represents temperature (in Kelvin), R is the ideal gas constant, and ΔG° is the standard Gibbs free energy change. The exercise shows how ΔG changes with different reaction conditions and requires calculating Q as an intermediate step to determine how far the system is from equilibrium.
Le Chatelier's Principle
Le Chatelier's principle is a qualitative tool that helps us predict how a system at equilibrium reacts to external changes. According to this principle, if a dynamic equilibrium is disturbed by altering conditions such as concentration, temperature, or pressure, the system will adjust itself in a way that counteracts the disturbance and re-establishes equilibrium.

For example, increasing the pressure by reducing the volume of a gaseous reaction would shift the equilibrium to the side with fewer moles of gas. Alternatively, if one of the reactants was added to the system, the reaction would shift towards the products to minimize the disturbance. Understanding Le Chatelier's principle enables us to control and predict the outcomes of reactions in a variety of scientific and industrial processes.
Standard State Equilibrium Constant (K)
The equilibrium constant, K, is a crucial numeric representation of the ratio of the concentrations of products to reactants at equilibrium in a reaction. Expressed in the standard state (usually 1 atm pressure and 298 K for gases and 1M concentration for solutions), K provides invaluable insight into the position of the equilibrium.

For reactions where ΔG° is known, the equilibrium constant can be calculated using the relationship:\[\begin{equation}K = \exp\left(-\frac{\Delta G^\circ}{RT}\right)\end{equation}\]

This equation highlights the inverse relationship between ΔG° and K. A negative ΔG° corresponds to a large K, indicating that at equilibrium, the reaction favors the formation of products. Conversely, a positive ΔG° leads to a small K, suggesting that the reactants are favored. This relationship is exemplified in the given exercise, which requires the computation of K using a provided ΔG° to then determine ΔG under non-standard conditions.

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Most popular questions from this chapter

Some nonelectrolyte solute (molar mass \(=142 \mathrm{~g} / \mathrm{mol}\) ) was dissolved in \(150 . \mathrm{mL}\) of a solvent \(\left(\right.\) density \(\left.=0.879 \mathrm{~g} / \mathrm{cm}^{3}\right)\). The elevated boiling point of the solution was \(355.4 \mathrm{~K}\). What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is \(33.90 \mathrm{~kJ} / \mathrm{mol}\), the entropy of vaporization is \(95.95 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\), and the boiling-point elevation constant is \(2.5 \mathrm{~K} \cdot \mathrm{kg} / \mathrm{mol}\).

Carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) form ideal solutions. Consider an equimolar solution of \(\mathrm{CCl}_{4}\) and \(\mathrm{C}_{6} \mathrm{H}_{6}\) at \(25^{\circ} \mathrm{C}\). The vapor above the solution is collected and condensed. Using the following data, determine the composition in mole fraction of the condensed vapor.

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ a. Without referring to Appendix 4 , predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperaturedependent. Predict the sign of \(\Delta S_{\text {surr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K}\). mol at \(298 \mathrm{~K}\). Using these values and data in Appendix 4 , calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C}\). The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .{ }^{\circ} \mathrm{C}\). f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\). The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\). g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\). Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\). The boiling point for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(29.0 \mathrm{~kJ} / \mathrm{mol}\). [Hint: The phase change reaction and the corresponding equilibrium expression are $$ \mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{Ni}(\mathrm{CO})} $$ \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) will liquefy when the pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is greater than the \(K\) value.]

Which of the following involve an increase in the entropy of the system? a. melting of a solid b. sublimation c. freezing d. mixing e. separation f. boiling

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

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