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

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

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: \[ Q_a = \frac{(50)^2}{200 \times (200)^3} \]

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: \[ Q_b = \frac{(200)^2}{200 \times (600)^3} \]

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: \[ K = \exp \left({-\frac{\Delta G^{\circ}}{RT}}\right) \]

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: \[ \Delta G = \Delta G^{\circ} + RT \ln Q_a \] \[ \Delta G_a = -33000\: \mathrm{J/mol} + (8.314\: \mathrm{J/(mol \: K)}) (298\: \mathrm{K}) \ln Q_a \]

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.

Key concepts

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.