Question

The following standard Gibbs energy changes are given for \(25^{\circ} \mathrm{C}\) (1) \(\mathrm{SO}_{2}(\mathrm{g})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=-246.4 \mathrm{kJ}\) (2) \(\mathrm{CS}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(\Delta G^{\circ}=-41.5 \mathrm{kJ}\) (3) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \longrightarrow \operatorname{COS}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=+1.4 \mathrm{kJ}\) (4) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\) \(\Delta G^{\circ}=-28.6 \mathrm{kJ}\) Combine the preceding equations, as necessary, to obtain \(\Delta G^{\circ}\) values for the following reactions. (a) \(\operatorname{COS}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow\) \(\begin{aligned} \mathrm{SO}_{2}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) & \Delta G^{\circ}=? \end{aligned}\) (b) \(\cos (g)+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow\) \(\mathrm{SO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \quad \Delta G^{\circ}=?\) \(\left.+\quad \mathrm{H}_{\mathrm{O}} \mathrm{C}(\mathrm{d})=\mathrm{CO}_{-}^{\circ} \mathrm{G}\right)+\mathrm{H}_{-}^{-} \mathrm{S}(\mathrm{q})\) (c) \(\cos (\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) \(\Delta G^{\circ}=?\) Of reactions (a), (b), and (c), which is spontaneous in the forward direction when reactants and products are present in their standard states?

Short answer

The standard Gibbs energies for the reactions are \( \Delta G^{\circ}(a) = -125 kJ, \Delta G^{\circ}(b) = -203.5 kJ, and \Delta G^{\circ}(c) = -70.1 kJ\). All three reactions (a), (b), and (c) are spontaneous under standard conditions.

Step-by-step solution

Reaction (a)

Using the principle of conservation of energy and composition, we write a reaction for (a) as a sum of the given reactions. Multiply reaction (1) by 1, reaction (2) by -1, reaction (3) by -1, and reaction (4) by 2. Add these resulting reactions up, and you will get the reaction for (a). The \( \Delta G^{\circ}\) value for the reaction results from the sum of the products of each reaction’s \( \Delta G^{\circ}\) and the respective multipliers. Thus, \( \Delta G^{\circ}(a) = 1* (-246.4 kJ) + -1*(-41.5 kJ) + -1*(1.4 kJ) + 2*(-28.6 kJ) = -125 kJ\)

Reaction (b)

Again, we write a reaction for (b) as a sum of the given reactions. Multiply reaction (1) by 1, reaction (2) by -1, and reaction (3) by -1. Summing up, we get the reaction for (b) and its standard Gibbs energy. Thus, \( \Delta G^{\circ}(b) = 1*(-246.4 kJ) + -1*(-41.5 kJ) + -1*(1.4 kJ) = -203.5 kJ\)

Reaction (c)

Now, we compose reaction (c) by summing up suitable combinations of the given reactions. Add reaction (2) and reaction (4) to get the reaction (c). As before, calculate \( \Delta G^{\circ}(c) \) as the sum of standard Gibbs energies of reactions (2) and (4), so \( \Delta G^{\circ}(c) = -41.5 kJ + -28.6 kJ = -70.1 kJ\)

Determine spontaneity

Finally, we can determine spontaneous reactions. Whenever \( \Delta G^{\circ}\) is negative, the reaction under standard conditions is spontaneous. In this case, all three reactions (a), (b), and (c) have negative \( \Delta G^{\circ}\) values, and thus all three are spontaneous under standard conditions.

Key concepts

Spontaneity of Reactions

Gibbs Free Energy is a cornerstone in understanding the spontaneity of chemical reactions. Spontaneity refers to whether a reaction will proceed on its own without external influence once it is initiated. To determine if a reaction is spontaneous under standard conditions, we examine the sign of the Gibbs Free Energy change, denoted as \( \Delta G^{\circ} \).

If \( \Delta G^{\circ} \) is negative, the reaction is considered spontaneous. This means that the reaction can occur without additional energy input at standard conditions, meaning a temperature of 25°C (298 K) and pressure of 1 atm for gases. Conversely, if \( \Delta G^{\circ} \) is positive, the reaction is non-spontaneous under these conditions.

Reactants and products must be in their standard states for this assessment to be valid. Spontaneous reactions under standard state conditions often release energy, contributing to the stability of the products formed. This concept is vital in predicting reaction behaviors and is used extensively in both academic and industrial chemistry.

Standard State Conditions

Standard state conditions are essential for expressing thermochemical equations uniformly. They provide a reference point, making thermodynamic data comparable across different reactions. These conditions are set at a temperature of 25°C (298 K) and 1 atmosphere pressure.

For gases, the standard state condition is a partial pressure of 1 atm. For solutions, it’s a concentration of 1 Molarity. For pure substances, the most stable physical form at 1 atm is used.

Working under standard conditions allows chemists to predict how a reaction will behave in a controlled setting and simplifies calculations of thermodynamic properties such as enthalpy, entropy, and Gibbs Free Energy. This uniformity is crucial in chemical thermodynamics and provides consistency in reporting and interpreting scientific data.

Thermochemistry

Thermochemistry is all about the study of heat change in chemical reactions. It deals with thermodynamic principles to assess the energy changes, specifically the heat involved, in a reaction.

The heat change associated with chemical reactions is characterized by δH, or the change in enthalpy. Reactions can either be exothermic, releasing heat, or endothermic, absorbing heat. In the context of Gibbs Free Energy, the relation \( \Delta G = \Delta H - T \Delta S \) connects enthalpy \( \Delta H \), temperature \( T \), and entropy \( \Delta S \).

Knowing the direction and magnitude of these energy changes is significant because it impacts whether a reaction is feasible or spontaneous. Thermochemistry helps in understanding reaction energetics and is valuable in industries such as fuels, materials production, and food processing. It is vital to thoroughly grasp this concept to predict how systems will behave under different conditions and to optimize reactions for desired outcomes.