Problem 111 - Chemistry An Atoms First Approach

Question

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

Step-by-Step Solution

Answer

The equilibrium constant for this reaction is 0.1088, and the equilibrium pressures of A and B are 0.9019 atm and 1.0981 atm, respectively. The calculations show that ΔG is close to 0 at the equilibrium conditions given, confirming that the system is at equilibrium.

Step 1: Calculate Gibbs free energy change

To find the equilibrium constant, we need to first find the Gibbs free energy change for the reaction, which is given by: \(\Delta G^{\circ} = G^{\circ}_{B} - G^{\circ}_{A}\). \( \Delta G^{\circ} = 11,718 \mathrm{J}/\mathrm{mol} - 8,996 \mathrm{J}/\mathrm{mol} = 2,722 \mathrm{J}/\mathrm{mol} \)

Step 2: Calculate the equilibrium constant K

The relationship between the Gibbs free energy change and the equilibrium constant K is given by: \(\Delta G^{\circ} = -RT\ln{K}\), where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin. We need to find the temperature in Kelvin first: \(T(K) = 25^{\circ}C + 273.15 = 298.15 K\) Now, we can rearrange the equation to solve for K: \(K = \exp{(-\frac{\Delta G^{\circ}}{RT})} \) \(K = \exp{(-\frac{2,722 \mathrm{J}/\mathrm{mol}}{8.314 \mathrm{J}/(\mathrm{mol}\cdot\mathrm{K})\cdot 298.15\mathrm{K}})} = 0.1088\) #a. Conclusion:# The equilibrium constant for this reaction is 0.1088. #b. Calculate the equilibrium pressures#

Step 3: Set up the reaction equation

Let's denote the initial pressures of A and B as \(P_A^0 = 1.00 \, \mathrm{atm}\) and \(P_B^0 = 1.00 \, \mathrm{atm}\), respectively. At equilibrium, the change in pressure will be given by: \(P_A^{eq} = P_A^0 - x\) and \(P_B^{eq} = P_B^0 + x\), where x is the change in pressure due to the reaction.

Step 4: Use equilibrium constant expression

The reaction has the form: \(A \rightleftharpoons B\), and the equilibrium constant expression will be: \(K = \frac{P_B^{eq}}{P_A^{eq}}\) Substitute the equilibrium pressures in the expression: \(0.1088 = \frac{1.00 \, \mathrm{atm} + x}{1.00 \, \mathrm{atm} - x}\)

Step 5: Solve for x

Now we need to find the value of x that brings the expression to equilibrium. Rearrange the equation to isolate x: \(x = \frac{1.00 \, \mathrm{atm} - 0.1088 \cdot (1.00 \, \mathrm{atm} - x)}{1 + 0.1088}\) Solve for x: \(x = 0.0981 \, \mathrm{atm}\)

Step 6: Calculate equilibrium pressures

Now that we have the value of x, we can find the equilibrium pressures for A and B: \(P_A^{eq} = 1.00 \, \mathrm{atm} - 0.0981 \, \mathrm{atm} = 0.9019 \, \mathrm{atm}\) \(P_B^{eq} = 1.00 \, \mathrm{atm} + 0.0981 \, \mathrm{atm} = 1.0981 \, \mathrm{atm}\) #b. Conclusion#: The equilibrium pressures of A and B are 0.9019 atm and 1.0981 atm, respectively. #c. Show that ΔG = 0 at equilibrium#

Step 7: Define the Gibbs free energy equation

The Gibbs free energy change equation is given by: \(\Delta G = \Delta G^{\circ} + RT\ln{Q}\), where \(Q\) is the reaction quotient which equals to \( \frac{P_B^{eq}}{P_A^{eq}} \) for this reaction.

Step 8: Calculate Q

Next, we need to calculate the reaction quotient, Q: \(Q = \frac{P_B^{eq}}{P_A^{eq}} = \frac{1.0981 \, \mathrm{atm}}{0.9019 \, \mathrm{atm}} = 1.2174\)

Step 9: Calculate ΔG

Plug in the calculated values to find ΔG: \(\Delta G = 2,722 \mathrm{J}/\mathrm{mol} + 8.314 \mathrm{J}/(\mathrm{mol}\cdot\mathrm{K}) \cdot 298.15\mathrm{K} \cdot \ln{1.2174} = -0.0006 \, \mathrm{J}/\mathrm{mol}\) The calculated ΔG is close to zero, which means that the system is already at equilibrium, considering the error in the numerical calculations. #c. Conclusion#: The calculations show that ΔG is close to 0 at the equilibrium conditions given, confirming that the system is at equilibrium.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy is a fundamental concept in chemistry that determines the spontaneity of a reaction. It combines enthalpy (the total energy of the system) and entropy (the measure of disorder or randomness). The formula for Gibbs Free Energy is: \[\Delta G = \Delta H - T\Delta S\]Where:

• \( \Delta G \) is the Gibbs Free Energy change.
• \( \Delta H \) is the change in enthalpy.
• \( T \) is the temperature in Kelvin.
• \( \Delta S \) is the change in entropy.

When \( \Delta G \) is negative, the reaction is spontaneous and can occur without external energy. When \( \Delta G \) is positive, the reaction is non-spontaneous and will not occur without energy input. At equilibrium, \( \Delta G = 0 \), meaning the system is stable and there is no net change in the concentrations of reactants and products. This concept is crucial in predicting the direction and extent of chemical reactions, ensuring no surprises during experiments.

Equilibrium Constant

The equilibrium constant, represented as \( K \), is an essential parameter in chemistry. It provides insight into the position of equilibrium for reversible reactions. The equilibrium constant arises from the equilibrium expression, which is based on the concentrations of products and reactants at equilibrium: \[K = \frac{[Products]^{coefficients}}{[Reactants]^{coefficients}}\]A large value of \( K \) indicates that at equilibrium, products predominate. Conversely, a small \( K \) suggests that reactants are more plentiful. Importantly, the equilibrium constant only changes with temperature, reflecting how dynamic equilibrium is sensitive to environmental conditions. The connection between \( \Delta G^{\circ} \) and \( K \) is pivotal: \[ \Delta G^{\circ} = -RT \ln K \] Where:

• \( R \) is the gas constant.
• \( T \) is the temperature in Kelvin.

This equation showcases the balance between energy and order guiding reactions toward equilibrium, providing insights into reaction feasibility and extent.

Reaction Quotient

The reaction quotient, denoted by \( Q \), is closely related to the equilibrium constant but applies to any point in a reaction, not just equilibrium. It uses the same form as the equilibrium constant:\[Q = \frac{[Products]^{coefficients}}{[Reactants]^{coefficients}}\]By comparing \( Q \) with \( K \), the direction of a reaction can be predicted:

• If \( Q < K \), the forward reaction is favored, pushing more reactants to form products.
• If \( Q > K \), the reverse reaction is favored, leading to the conversion of products back to reactants.
• If \( Q = K \), the system is at equilibrium, and no net change occurs.

This contextual understanding of \( Q \) helps chemists control reaction conditions and substances' concentrations to achieve desired outcomes, boosting efficiency and safety in chemical processes.