Problem 61 - Chemistry: The Central Science

Question

Using data from Appendix \(C\), calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions. (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow 3 \mathrm{NO}(g)\) (c) \(6 \mathrm{Cl}_{2}(g)+2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{FeCl}_{3}(s)+3 \mathrm{O}_{2}(g)\) (d) \(\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

Step-by-Step Solution

Answer

(a) \(\Delta G^{\circ} = -140 \: kJ/mol\); spontaneous (b) \(\Delta G^{\circ} = 104.3 \: kJ/mol\); non-spontaneous (c) \(\Delta G^{\circ} = -117.8 \: kJ/mol\); spontaneous (d) \(\Delta G^{\circ} = -156.8 \: kJ/mol\); spontaneous

Step 1: Write down the ∆G° values from Appendix C

For this reaction, we need the standard Gibbs free energy change of formation (∆Gf°) for each reactant and product: \(SO_2(g) : -300.4 \hspace{3mm} kJ/mol\) \(O_2(g) : 0 \hspace{3mm} kJ/mol\) (by definition) \(SO_3(g) : -370.4 \hspace{3mm} kJ/mol\)

Step 2: Calculate ∆G° for the reaction

Apply the formula ∆G° = ∆G°(product) - ∆G°(reactant). For this reaction: ∆G° = \(2(-370.4) - (2(-300.4) + 0)\) ∆G° = \((-740.8) - (-600.8))\) ∆G° = \(-140\; kJ/mol\) Since ∆G° is negative, the reaction is spontaneous under standard conditions at 298 K. (b) \(NO_2(g) + N_2O(g) \longrightarrow 3 NO(g)\)

Step 1: Write down the ∆G° values from Appendix C

We need the ∆Gf° for each reactant and product: \(NO_2(g) : 51.3 \hspace{3mm} kJ/mol\) \(N_2O(g) : 104.2 \hspace{3mm} kJ/mol\) \(NO(g) : 86.6 \hspace{3mm} kJ/mol\)

Step 2: Calculate ∆G° for the reaction

Apply the formula: ∆G° = \(3(86.6) - (51.3 + 104.2)\) ∆G° = \(259.8 - 155.5\) ∆G° = \(104.3\; kJ/mol\) Since ∆G° is positive, the reaction is non-spontaneous under standard conditions at 298 K. (c) \(6 Cl_2(g) + 2 Fe_2O_3(s) \longrightarrow 4 FeCl_3(s) + 3 O_2(g)\)

Step 1: Write down the ∆G° values from Appendix C

We need the ∆Gf° for each reactant and product: \(Cl_2(g) : 0 \hspace{3mm} kJ/mol\) (by definition) \(Fe_2O_3(s) : -740.5 \hspace{3mm} kJ/mol\) \(FeCl_3(s) : -399.7 \hspace{3mm} kJ/mol\) \(O_2(g) : 0 \hspace{3mm} kJ/mol\) (by definition)

Step 2: Calculate ∆G° for the reaction

Apply the formula: ∆G° = \((4(-399.7) + 3(0)) - (6(0) + 2(-740.5))\) ∆G° = \((-1598.8) - (-1481.0))\) ∆G° = \(-117.8\; kJ/mol\) Since ∆G° is negative, the reaction is spontaneous under standard conditions at 298 K. (d) \(SO_2(g) + 2 H_2(g) \longrightarrow S(s) + 2 H_2O(g)\)

Step 1: Write down the ∆G° values from Appendix C

We need the ∆Gf° for each reactant and product: \(SO_2(g) : -300.4 \hspace{3mm} kJ/mol\) \(H_2(g) : 0 \hspace{3mm} kJ/mol\) (by definition) \(S(s) : 0 \hspace{3mm} kJ/mol\) (by definition) \(H_2O(g) : -228.6 \hspace{3mm} kJ/mol\)

Step 2: Calculate ∆G° for the reaction

Apply the formula: ∆G° = \((0 + 2(-228.6)) - (-300.4 + 2(0))\) ∆G° = \((-457.2) - (-300.4))\) ∆G° = \(-156.8\; kJ/mol\) Since ∆G° is negative, the reaction is spontaneous under standard conditions at 298 K.

Key Concepts

Spontaneous Reactions

Understanding spontaneous reactions is critical for students studying chemistry. A spontaneous reaction is one that proceeds on its own without any continuous external influence, such as the addition of energy. Instead, these reactions release energy, which is often in the form of heat and may result in a decrease in enthalpy (H), an increase in entropy (S), or both.

Importantly, the spontaneity of a reaction is determined by the Gibbs Free Energy change (G), which combines both enthalpy and entropy into one value. The general rule is that if G is negative (G < 0), the reaction will proceed spontaneously under the given conditions. Conversely, a positive G value (G > 0) means that the reaction is not spontaneous and will require energy input to occur. This helps students to quickly assess the spontaneity of reactions by just looking at the sign of G.

Standard Gibbs Free Energy Change

When discussing thermodynamics in chemistry, the concept of standard Gibbs free energy change (G°) is a pivotal one. This is an essential parameter representing the free energy change of a reaction carried out at standard conditions, which typically mean a temperature of 298 K (approximately 25 °C), a pressure of 1 atm, and all substances in their standard states.

Calculating G° requires knowledge of the G° values of individual reactants and products, which can often be found in a chemical reference appendix, and following the formula G = G°(products) - G°(reactants). A negative G° value suggests the reaction should occur spontaneously under standard conditions, while a positive value suggests it requires an energy input to proceed.

Thermodynamics in Chemistry

Thermodynamics in chemistry is the study of energy changes accompanying chemical transformations. This field of study provides the framework for predicting whether a chemical reaction will proceed and if so, to what extent. The laws of thermodynamics govern these energy changes and their implications for reaction spontaneity.

The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transferred or transformed. This means that the energy released or absorbed in a chemical reaction must be accounted for, typically as heat or work. The second law of thermodynamics introduces entropy, stating that the total entropy of an isolated system can never decrease over time. These foundational principles help students understand the driving forces behind chemical reactions and how energy is conserved and transferred in the process.

Chemical Reaction Spontaneity

Chemical reaction spontaneity is of great interest to students learning about chemical reactions. Spontaneity does not refer to the speed of a reaction but to the tendency of a reaction to occur without being driven by external energy. It's a common misconception that spontaneous reactions happen quickly, while in fact, some can be very slow.

The measurement of spontaneity is largely based on Gibbs free energy. Factors like temperature, pressure, and concentration can all influence reaction spontaneity, which is why G is calculated under specified conditions. Students are often taught to determine spontaneity by calculating the G value of a reaction, but understanding that reaction conditions can shift spontaneity is equally important. This ties into equilibrium and Le Chatelier's principle, where changes to reaction conditions can shift the position of equilibrium, affecting the spontaneity of the forward and reverse reactions.