Question

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

In summary, for the given reaction under the specified conditions, we find that: - For condition (a) with \(T=298\,K\), \(P_{N_2}=P_{H_2}=200\,atm\), and \(P_{NH_3}=50\,atm\), the change in Gibbs Free Energy (ΔG) is \(-24.0\,kJ/mol\). - For condition (b) with \(T=298\,K\), \(P_{N_2}=200\,atm\), \(P_{H_2}=600\,atm\), and \(P_{NH_3}=200\,atm\), the change in Gibbs Free Energy (ΔG) is \(-25.4\,kJ/mol\).

Step-by-step solution

Write the expression for the reaction quotient Q

The balanced chemical equation for the given reaction is: \[N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g)\] The reaction quotient (Q) can be found using the following formula, which includes the pressure of the gaseous substances: \[Q = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3}\]

Condition (a): T=298K, P_N2=200 atm, P_H2= 200 atm, P_NH3=50 atm

Calculate the value of Q

Using the given pressures for condition (a): \[Q_a = \frac{(50)^2}{(200)(200)^3} = 6.25 \times 10^{-9}\]

Calculate ΔG° from the standard Gibbs free energies of formation

We will use the standard Gibbs free energies of formation (in kJ/mol) for each gas as follows: \[\Delta G_{f} ^{\circ} (N_2) = 0\] \[\Delta G_{f} ^{\circ} (H_2) = 0\] \[\Delta G_{f} ^{\circ} (NH_3) = -16.7\] Now, calculate ΔG° for the reaction using: \[\Delta G^{\circ} = 2 \cdot (-16.7) - (1 \cdot 0 + 3 \cdot 0) = -33.4\, kJ/mol\]

Calculate ΔG for the reaction

Use the calculated values of ΔG° and Q to find ΔG: \[\Delta G= \Delta G^{\circ} + RT \ln{Q}\] For condition (a), plug in the values: \[\Delta G_a = -33.4\, kJ/mol + (8.314 \times 10^{-3}\, kJ/mol\cdot K)(298\, K) \ln(6.25 \times 10^{-9}) = -24.0\, kJ/mol\] Now, follow the same steps for condition (b).

Condition (b): T=298K, P_N2=200 atm, P_H2= 600 atm, P_NH3=200 atm

Calculate the value of Q

Using the given pressures for condition (b): \[Q_b = \frac{(200)^2}{(200)(600)^3} = 1.85 \times 10^{-9}\]

Calculate ΔG for the reaction

Use the calculated values of ΔG° and Q to find ΔG: \[\Delta G= \Delta G^{\circ} + RT \ln{Q}\] For condition (b), plug in the values: \[\Delta G_b = -33.4\, kJ/mol + (8.314 \times 10^{-3}\, kJ/mol\cdot K)(298\, K) \ln(1.85 \times 10^{-9}) = -25.4\, kJ/mol\] To summarize the results: Condition (a): ΔG = -24.0 kJ/mol Condition (b): ΔG = -25.4 kJ/mol

Key concepts

Gibbs Free Energy

Understanding the concept of Gibbs free energy, commonly denoted as \( \Delta G \), is crucial in the study of chemical thermodynamics. Simply put, it's the measure of the maximum reversible work a thermodynamic system can perform at constant temperature and pressure. This unique energy represents the balance between enthalpy (total heat content) and entropy (measure of disorder) in a given reaction. The relationship is expressed in the formula \( \Delta G = \Delta H - T\Delta S \), where \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.

When \( \Delta G \) is negative, the process or reaction occurs spontaneously, indicating that it's energetically favorable. However, if \( \Delta G \) is positive, external energy is required for the process to occur. In a situation where \( \Delta G \) is zero, the system is said to be at equilibrium, meaning the forward and reverse reactions occur at the same rate. This understanding of \( \Delta G \) can help you determine the likelihood of a reaction under specific conditions.

Reaction Quotient (Q)

The reaction quotient, denoted as \( Q \), is an expression that provides a snapshot of a reaction's position at any point in time. It's calculated using the same formula as the equilibrium constant (K), but with the current concentrations or pressures of the reactants and products, rather than their equilibrium values. The general form for a reaction quotient in a gaseous system is \( Q = \frac{(P_{products})}{(P_{reactants})} \), where \( P \) stands for partial pressure. For the given habitation reaction, \( Q \) is calculated as \( \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \).

Comparing the value of \( Q \) to the equilibrium constant \( K \) gives insight into which direction the reaction will proceed to achieve equilibrium. If \( Q < K \) the reaction will proceed forward, making more products. If \( Q > K \) the reaction will proceed in the reverse direction, making more reactants. When \( Q = K \) the system is at equilibrium, and no net reaction occurs.

Le Chatelier's Principle

Le Chatelier's principle is a fundamental concept in chemistry that predicts how a system at equilibrium responds to changes in concentration, pressure, or temperature. It states that if an external stress is applied to a system in dynamic equilibrium, the system will adjust its position to counteract the change and restore equilibrium.

For example, if the pressure is increased for a reaction where gaseous reactants and products are involved, the system will shift towards the side with fewer moles of gas to reduce the pressure. Similarly, if the temperature is increased for an exothermic reaction, the system will favor the forward reaction to absorb excess heat, shifting the equilibrium to the left and producing more reactants. This principle allows chemists to understand and control the direction of chemical reactions, optimizing the yield of desired products.

Standard Gibbs Free Energies of Formation

The standard Gibbs free energies of formation, \( \Delta G_f^\circ \), provide the values for the change in Gibbs free energy when one mole of a compound is formed from its elements under standard conditions (1 atm pressure, 25°C or 298 K). For pure elements in their standard states, such as \( N_2(g) \) and \( H_2(g) \) in their diatomic forms, the standard Gibbs free energy of formation is zero because there is no change during formation. However, for a compound like ammonia, \( NH_3(g) \) the value is not zero (in this case, -16.7 kJ/mol) and reflects the energy change during its formation.

The standard Gibbs free energies of formation are essential for calculating the standard Gibbs free energy change (\( \Delta G^\circ \) of a reaction. By applying Hess's law, we can sum the \( \Delta G_f^\circ \) values of the products and subtract those of the reactants, enabling us to predict the spontaneity of a reaction under standard conditions. This calculation plays a crucial role in thermochemical equations and helps in understanding the energetics of chemical reactions.