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Chapter 16: Problem 57

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

Short Answer

Expert verified
The standard Gibbs free energy change for the target reaction, \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\), is \(\Delta G^{\circ}=-1291 \mathrm{kJ}\).

Step by step solution

01

Write down the target reaction

Write down the target reaction: $$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$
02

Reverse Reaction 1

Reverse Reaction 1 to get \(\mathrm{CH}_{4}(g)\) on the reactant side: $$\mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \quad \Delta G_{1}^{\circ}=51 \mathrm{kJ}$$
03

Double Reaction 2

Double Reaction 2 to get \(2 \mathrm{O}_{2}(g)\) and \(2 \mathrm{H}_{2}\mathrm{O}(l)\) on the product side: $$4 \mathrm{H}_{2}(g)+2\mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{H}_{2}\mathrm{O}(l) \quad \Delta G_{2}^{\circ}=-2 \times 474 \mathrm{kJ}=-948 \mathrm{kJ}$$
04

Add the manipulated reactions

Add the manipulated Reaction 1 and the double of Reaction 2, with Reaction 3. This will give the target reaction: $$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ Now, sum the corresponding Gibbs free energy changes: $$\Delta G_{total}^{\circ} = \Delta G_{1}^{\circ} + \Delta G_{2}^{\circ} + \Delta G^{\circ}_3 = 51 \mathrm{kJ} + (-948 \mathrm{kJ}) + (-394 \mathrm{kJ}) = -1291 \mathrm{kJ}$$
05

Report the result

The standard Gibbs free energy change for the target reaction is: $$\Delta G^{\circ}(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l))=-1291 \mathrm{kJ}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
When looking at the chemical reaction given in the exercise, it's essential to understand what happens at the molecular level. The target reaction is - \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\).Chemical reactions involve breaking bonds in reactants and forming new bonds in products. During this process, bonds in methane \(\mathrm{CH}_{4}\) and oxygen \(\mathrm{O}_{2}\) are broken and new bonds are formed to create carbon dioxide \(\mathrm{CO}_{2}\) and water \(\mathrm{H}_{2}\mathrm{O}\). Each of these steps requires energy and involves rearranging atoms to make the products. - Reactants: Substances that undergo change (\(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\)).- Products: New substances formed (\(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\)).Chemical reactions are a vital part of thermodynamics, helping us understand the balance of energy in reactions. Here, we focus on Gibbs Free Energy, which tells us if a reaction is spontaneous or if it requires external intervention.
Enthalpy
Enthalpy is a concept that captures the total heat content of a system during a chemical reaction. While it is related to Gibbs Free Energy, it is not the same. Enthalpy is concerned with the heat absorbed or released at constant pressure.- Endothermic reactions absorb heat.- Exothermic reactions release heat.In the context of the reaction \(\mathrm{CH}_{4}+2\mathrm{O}_{2}\rightarrow\mathrm{CO}_{2}+2\mathrm{H}_{2}\mathrm{O}\), the reaction is known to be exothermic. It releases energy in the form of heat. This aligns with the calculation showing a negative Gibbs Free Energy (\(-1291 \mathrm{kJ}\)) confirming the spontaneity of the reaction. Enthalpy changes, along with entropy changes, allow us to calculate Gibbs Free Energy and decide if a reaction will proceed on its own. Understanding enthalpy helps predict reaction behavior and energy changes when substances react.
Thermodynamics
Thermodynamics is the study of energy and its transformations. In a chemical reaction, it helps us predict if a reaction will take place based on energy changes. Three main components contribute to this:1. **Enthalpy (\(H\)):** - Deals with heat changes at constant pressure.2. **Entropy (\(S\)):** - A measure of disorder or randomness in a system. - Higher entropy means more disorder.3. **Gibbs Free Energy (\(G\)):** - Combines enthalpy and entropy to determine reaction spontaneity. - Calculated by the formula: \[\Delta G = \Delta H - T\Delta S\]- **Negative \(\Delta G\):** Spontaneous reaction.- **Positive \(\Delta G\):** Non-spontaneous reaction.In the reaction analyzed, the negative \(-1291 \mathrm{kJ}\) indicates that it is spontaneous under standard conditions. Thermodynamics provides a framework to understand how energy changes in reactions, particularly through Gibbs Free Energy, assisting in predicting whether reactions need energy input or occur naturally.

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Most popular questions from this chapter

For the reaction at \(298 \mathrm{K}\), $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{kJ}\) and \(-176.6 \mathrm{J} / \mathrm{K},\) respectively. What is the value of \(\Delta G^{\circ}\) at 298 K? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

Consider the reaction: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ for which \(\Delta H\) is \(-233 \mathrm{kJ}\) and \(\Delta S\) is \(-424 \mathrm{J} / \mathrm{K}.\) a. Calculate the free energy change for the reaction \((\Delta G)\) at \(393 \mathrm{K}.\) b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

In the text, the equation $$\Delta G=\Delta G^{\circ}+R T \ln (Q)$$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of mol/L for the quantities in \(Q,\) specifically for aqueous reactions. With this in mind, consider the reaction $$HF(a q) \rightleftharpoons H^{+}(a q)+F^{-}(a q)$$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C}.\) a. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 M\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} M\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} M\) Based on the calculated \(\Delta G\) values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{kJ} / \mathrm{mol} .\) In a particular experiment, equal moles of \(\mathrm{H}_{2}(g)\) at 1.00 atm and \(\mathrm{Br}_{2}(g)\) at 1.00 atm were mixed in a 1.00-L flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ},\) and \(\Delta S^{\circ} .\)

At \(1500 \mathrm{K},\) the process $$\begin{aligned} &\mathbf{I}_{2}(g) \longrightarrow 2 \mathbf{I}(g)\\\ &10 \mathrm{atm} \quad 10 \mathrm{atm} \end{aligned}$$ is not spontaneous. However, the process $$\begin{aligned} &\mathbf{I}_{2}(g) \longrightarrow 2 \mathbf{I}(g)\\\ &0.10 \mathrm{atm} \quad 0.10 \mathrm{atm} \end{aligned}$$ is spontaneous at \(1500 \mathrm{K}\). Explain.
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