Question

Given the following data: $$\begin{aligned}2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \longrightarrow \mathrm{CH}_{4}(g) & & \Delta G^{\circ}=-51 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) & \Delta \mathrm{H}_{2} \mathrm{O}(l) & & \Delta G^{\circ}=-474 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ} &=-394 \mathrm{~kJ}\end{aligned}$$ Calculate \(\Delta G^{\circ}\) for \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(l) .\)

Short answer

The standard Gibbs free energy change for the target reaction, \( \mathrm{CH}_{4}(g)+2\ \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2\ \mathrm{H}_{2}O(l) \), is \(\Delta G^{\circ} = -131\ \mathrm{kJ}\).

Step-by-step solution

Identify the target reaction

The target reaction for which we have to find the standard Gibbs free energy change is: \( \mathrm{CH}_{4}(g)+2\ \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2\ \mathrm{H}_{2}O(l) \)

Combine given reactions to match the target reaction

We need to find a way to combine the given reactions in a way that will result in the target reaction. After examination, we can see that adding the 1st and 2nd reactions, and then the 3rd reaction in reverse, will resemble our target reaction: (1) \( 2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \longrightarrow \mathrm{CH}_{4}(g) \) (2) \( 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2}O(l) \) (3, reverse) \( \mathrm{CO}_{2}(g) \longrightarrow \mathrm{C}(s)+\mathrm{O}_{2}(g) \) >equals> \( \mathrm{CH}_{4}(g) + 2\ \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + 2\ \mathrm{H}_{2}O(l) \)

Calculate \(\Delta G^{\circ}\) for the target reaction

To find the standard Gibbs free energy change for the target reaction, we will sum the respective \(\Delta G^{\circ}\) values of the reactions we used in Step 2, considering the appropriate signs: \(\Delta G^{\circ}_{\text{target}} = \Delta G^{\circ}_{1} + \Delta G^{\circ}_{2} - \Delta G^{\circ}_{3}\) (Note that we subtracted \(\Delta G^{\circ}_{3}\) since we used the reverse form of reaction 3) Now, plug in the given values: \(\Delta G^{\circ}_{\text{target}} = (-51\ \mathrm{kJ}) + (-474\ \mathrm{kJ}) - (-394\ \mathrm{kJ})\) \(\Delta G^{\circ}_{\text{target}} = -51\ \mathrm{kJ} - 474\ \mathrm{kJ} + 394\ \mathrm{kJ}\) \(\Delta G^{\circ}_{\text{target}} = -131\ \mathrm{kJ}\) Thus, the standard Gibbs free energy change for the target reaction is \(\Delta G^{\circ} = -131\ \mathrm{kJ}\).

Key concepts

Thermodynamics

Thermodynamics is a fundamental principle in chemistry that deals with the study of energy transformations. One key aspect is Gibbs Free Energy, which helps predict the feasibility of a chemical reaction under constant temperature and pressure. Gibbs Free Energy (\( \Delta G \)) is defined by the equation:\[\Delta G = \Delta H - T\Delta S\]

• Where \( \Delta G \) is the change in free energy
• \( \Delta H \) is the change in enthalpy (heat content)
• \( T \) is the temperature in Kelvin
• \( \Delta S \) is the change in entropy (disorder)

The sign of \( \Delta G \) is crucial:- A negative \( \Delta G \) indicates a spontaneous reaction, meaning it can occur without external input.- A positive \( \Delta G \) suggests the reaction is non-spontaneous and needs energy input to proceed.
This concept is crucial in understanding why certain processes occur naturally. In the context of the exercise, we calculated \( \Delta G^\circ\) for a specified chemical reaction, determining it was spontaneous under standard conditions because \( \Delta G^\circ\) was negative.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of bonds, resulting in new products. Understanding how Gibbs Free Energy relates to reaction spontaneity is key to predicting and controlling these reactions. In the exercise, we began by identifying the target chemical reaction: \[ \mathrm{CH}_4(g) + 2 \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + 2 \mathrm{H}_2O(l) \]To solve Gibbs Free Energy problems like this one, a systematic approach of combining known reactions is used. This exercise highlighted the importance of manipulating reactions to match the desired equation through techniques like reversing reactions and summing \( \Delta G^\circ \) values. Each adjustment to a chemical equation must consider change in free energy, emphasizing a central role thermodynamics plays in chemistry. Ultimately, these techniques assist chemists in determining whether a reaction will proceed under specified conditions.

Standard Conditions

Standard conditions provide a set reference point in thermodynamics. When a process is analyzed, it is often declared to occur under 'standard conditions,' which means:

• The temperature is at 298 K (25°C)
• The pressure is at 1 atm
• Substances are in their standard states, which means the most stable form of the substance at 1 atm and at the specified temperature

Calculations utilizing standard Gibbs Free Energy change (\(\Delta G^\circ\)) are made under these conditions. This allows chemists to compare different reactions uniformly. In the exercise, solving for \(\Delta G^\circ \) allowed us to determine deterministically the spontaneity of the chemical reaction at standard conditions. Importantly, deviations from this set point in real-world applications might change the reaction dynamics. Therefore, while standard conditions give a baseline for theoretical calculations, actual laboratory conditions could vary, and such changes need to be factored into comprehensive chemical analysis.