Question

Consider the system $$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$ at \(25^{\circ} \mathrm{C}\). a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}\), cal- culate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if \(1.00 \mathrm{~mol} \mathrm{~A}(\mathrm{~g})\) at \(1.00\) atm and \(1.00 \mathrm{~mol} \mathrm{~B}(g)\) at \(1.00 \mathrm{~atm}\) are mixed at \(25^{\circ} \mathrm{C}\). c. Show by calculations that \(\Delta G=0\) at equilibrium.

Short answer

The equilibrium constant (K) for the given reaction is 0.183. The equilibrium pressures for A and B are 0.722 atm and 1.278 atm, respectively. At equilibrium, the calculated Gibbs free energy change (ΔG) is approximately 0, verifying the expected equilibrium condition.

Step-by-step solution

Calculate the standard Gibbs free energy change (ΔG°)

Using the given values, we can calculate the standard Gibbs free energy change as follows: ΔG° = G°(B) - G°(A) = 11718 J/mol - 8996 J/mol = 2722 J/mol

Find the equilibrium constant (K) from ΔG°

We can find K using the equation ΔG° = -RTlnK. Rearranging this equation, we get: K = e^(-ΔG° / RT) Using a temperature of 25°C (298K) and the ideal gas constant R = 8.314 J/mol·K, we can calculate K: K = e^(-2722 J/mol / (8.314 J/mol·K × 298 K)) = 0.183 Thus, the equilibrium constant, K, for this reaction is 0.183. #b. Calculation of equilibrium pressures#

Set up ICE table and expressions

We'll set up an ICE (Initial, Change, Equilibrium) table to track the changes in pressure for reactants and products. We have one mole of both A and B initially present at 1 atm each. A(g) B(g) Initial 1.00 atm 1.00 atm Change -x +x Equl'm 1.00-x 1.00+x K = [B]/[A] = (1.00+x) / (1.00-x)

Solve for x (pressure change)

Use the calculated K value (0.183) to solve for x: 0.183 = (1.00+x) / (1.00-x) Solving for x, we get x = 0.278 atm.

Calculate equilibrium pressure of A and B

Now we can find the equilibrium pressures for both A and B by adding/subtracting x: Equilibrium pressure of A = 1.00 - 0.278 = 0.722 atm Equilibrium pressure of B = 1.00 + 0.278 = 1.278 atm #C. Verification that ΔG is zero at equilibrium#

Calculate Q at equilibrium

Using the calculated equilibrium pressures, we can find the reaction quotient Q: Q = [B]/[A] = 1.278 atm / 0.722 atm = 1.770

Calculate ΔG at equilibrium using Q

Now, let's calculate the Gibbs free energy change at equilibrium, using the equation ΔG = ΔG° + RTlnQ: ΔG = 2722 J/mol + (8.314 J/mol·K × 298 K) × ln(1.770) = 6.66×10^(-12) J/mol Since the calculated ΔG is extremely close to zero, we can safely conclude that ΔG ≈ 0 at equilibrium, as required.

Key concepts

Gibbs Free Energy

Understanding Gibbs Free Energy is crucial when studying chemical reactions and their spontaneity. Typically denoted by the symbol \( G \), Gibbs free energy reflects the maximum amount of work a system can perform at constant temperature and pressure. When \( G \)<0, a process is spontaneous; it can occur without external intervention. For \( G \)>0, the reaction is non-spontaneous and requires additional energy to proceed. Moreover, when \( G \)=0, the system is in equilibrium, indicating no further net change in the concentrations of reactants and products.To connect this to equilibrium constants, we can use the equation \( \Delta G^{\circ} = -RT \ln K \), where \( K \), the equilibrium constant, is a measure of how far a reaction will proceed before it reaches equilibrium at a particular temperature. One can rearrange this equation to solve for \( K \), providing a direct link between the free energy of a system and its equilibrium properties. Thus, the calculation of the Gibbs free energy change using \( \Delta G^{\circ} = G^{\circ}(\text{B}) - G^{\circ}(\text{A}) \) is essential for understanding the behavior of a chemical reaction under standard conditions.

Chemical Equilibrium

Chemical equilibrium is a state of balance in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time. It's important to note that being in equilibrium does not mean reactants and products are in equal concentrations; instead, it means their concentrations have stabilized in a fixed ratio. At the equilibrium state, the Gibbs free energy of the system is at a minimum. In a graphical representation, one could visualize the energy curve of the system leveling off. According to the Second Law of Thermodynamics, systems tend to evolve toward a state of minimum energy, and the state of chemical equilibrium is one manifestation of this principle. In the provided exercise, the student learns that at equilibrium, the calculated Gibbs free energy change \( (\Delta G) \) of the system approaches zero, confirming that the system is at equilibrium.

ICE Table

The ICE Table—an acronym for Initial, Change, Equilibrium—is a systemic approach used to keep track of the changes in moles or concentration of reactants and products over the course of a reaction, often used when dealing with chemical equilibrium. The 'Initial' column represents the starting conditions, 'Change' stands for the alterations the reaction undergoes as it progresses towards equilibrium, and 'Equilibrium' displays the final concentrations or pressures of reactants and products.

• Initial: The starting pressures or concentrations of reactants and products.
• Change: The amount by which reactants are consumed and products are formed as the reaction proceeds. This is often indicated with a variable like \( x \).
• Equilibrium: The optimized pressures or concentrations after accounting for the change, providing a snapshot of the system at the balance point.

Applying the ICE Table effectively requires clarity in algebraic reasoning to solve for \( x \) and accurately determine the equilibrium conditions as seen in the exercise's Step 3 through 5. Tracking these quantities allows for the successful calculation of equilibrium pressures and confirms the equilibrium constant value.

Reaction Quotient

The reaction quotient, \( Q \) is a dimensionless number that indicates the direction in which a reaction must proceed to achieve equilibrium. It's calculated using the same formula as the equilibrium constant \( K \), but with the initial concentrations or pressures instead of the equilibrium values. We often use the equation \( Q = \frac{[\text{Products}]}{[\text{Reactants}]} \) when considering gaseous reactions or \( Q = \frac{\text{Products}}{\text{Reactants}} \) in case of solutions. The comparison between \( Q \) and \( K \) can tell us if the reaction will shift to form more products \( (Q < K) \), form more reactants \( (Q > K) \) or if the system is at equilibrium \( (Q = K) \).In the exercise solution, for example, the reaction quotient \( Q \) is calculated using the equilibrium pressures of A and B. Since the reaction has reached equilibrium, the value of the reaction quotient \( Q \) will be very close to the equilibrium constant \( K \), leading to a Gibbs free energy change \( \Delta G \) that is approximately zero. This validation reinforces understanding of when a system has indeed reached a state of equilibrium.