Problem 77 - Chemistry: The Central Science in SI Units, Global Edition

Question

Consider the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (a) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of all gases are \(33.4 \mathrm{kPa}\).

Step-by-Step Solution

Answer

(a) The standard Gibbs free energy change (∆G°) at 298 K is -171.2 kJ/mol. (b) The non-standard Gibbs free energy change (∆G) at 298 K with the given partial pressures is -142627.28 J/mol, which is approximately -142.63 kJ/mol.

Step 1: Calculate standard Gibbs free energy change

Using the ∆G° value for each species provided in Appendix C, we can calculate the standard Gibbs free energy change for the reaction. We have the reaction: 2 NO(g) + O₂(g) ⟶ 2 NO₂(g) Using the formula mentioned above, we get: ∆G° = Σ(ν_product * ∆G°_product) - Σ(ν_reactant * ∆G°_reactant) ∆G° = (2 * ∆G°(NO₂) - (2 * ∆G°(NO) + ∆G°(O₂)) Plug in the values for each species: ∆G° = (2 * 51.3 kJ/mol) - (2 * 86.6 kJ/mol + 0 kJ/mol) Calculate the Gibbs free energy change: ∆G° = -171.2 kJ/mol

Step 2: Calculate reaction quotient

We are given that the partial pressures of all gases are 33.4 kPa. Calculate the reaction quotient Q using the partial pressures: Q = [NO₂]² / ([NO]² * [O₂]) Since the partial pressures are the same for each gas (33.4 kPa), we can write: Q = (33.4)² / ((33.4)² * 33.4) Q = 1 / 33.4

Step 3: Calculate non-standard Gibbs Free Energy Change

Now, we can calculate the non-standard Gibbs free energy change (∆G) using the Van't Hoff equation: ∆G = ∆G° + RT ln(Q) Plug in the values: ∆G = -171.2 kJ/mol + (8.314 J/mol*K * 298 K * ln(1/33.4)) Don't forget to convert kJ to J: ∆G = -171200 J/mol + (8.314 J/mol*K * 298 K * ln(1/33.4)) Calculate the value to obtain the final answer: ∆G = -142627.28 J/mol In conclusion, (a) The standard Gibbs free energy change (∆G°) at 298 K is -171.2 kJ/mol. (b) The non-standard Gibbs free energy change (∆G) at 298 K with the given partial pressures is -142627.28 J/mol, which is approximately -142.63 kJ/mol.

Key Concepts

Chemical Thermodynamics

Chemical thermodynamics helps us understand how and why chemical reactions occur. At its core, it's about energy - tracking energy changes in reactions. Gibbs Free Energy (denoted as \(\Delta G\)) is a key part of this.

Gibbs Free Energy combines enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) to predict whether a reaction is spontaneous. It's defined by the equation:

• \[ \Delta G = \Delta H - T \Delta S \]

This equation tells us about the feasibility of reactions:

• If \(\Delta G\) is negative, the reaction is spontaneous and can proceed without added energy.
• If \(\Delta G\) is positive, the reaction isn't spontaneous and needs energy to proceed.

Calculating \(\Delta G\) involves standard free energy changes (\(\Delta G^\circ\)), which use values at standard conditions (25°C, 1 atm pressure). This helps us predict reaction behavior under these conditions.

Reaction Quotient

The reaction quotient \(Q\) is a snapshot of a reaction's position at any given moment. Calculating \(Q\) involves the concentrations or partial pressures of reactants and products at a specific time.

For the reaction \(2 \text{NO}(g) + \text{O}_2(g) \longrightarrow 2 \text{NO}_2(g)\), the reaction quotient \(Q\) is calculated by:

• \[ Q = \frac{[\text{NO}_2]^2}{[\text{NO}]^2 [\text{O}_2]} \]

When \(Q\) equals the equilibrium constant \(K\), the reaction is at equilibrium. If \(Q eq K\), the reaction will shift to reach equilibrium:

• If \(Q
• If \(Q>K\), the reaction favors the reverse direction (forming reactants).

In non-standard conditions, using \(Q\) helps calculate \(\Delta G\), which guides our predictions about the reaction's direction and spontaneity.

Non-Standard Conditions

Reactions often occur under non-standard conditions, with varying temperatures, pressures, or concentrations. These require adjustments to the standard Gibbs Free Energy, \(\Delta G^\circ\).

Non-standard Gibbs Free Energy (\(\Delta G\)) can be calculated using the expression:

• \[ \Delta G = \Delta G^\circ + RT \ln(Q) \]

Here, \(R\) is the universal gas constant, \(T\) is temperature in Kelvin, and \(Q\) is the reaction quotient.

This equation adjusts \(\Delta G^\circ\) to account for actual conditions. For our example under non-standard conditions, with specific partial pressures, \(\Delta G\) changes and tells us about the reaction's spontaneity in real life scenarios.

• A negative \(\Delta G\) indicates spontaneous reactions even at non-standard conditions.
• A positive \(\Delta G\) suggests the need for additional energy.

Understanding these adjustments allows chemists to predict and control chemical reactions effectively, even when conditions stray from the standard.