Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 64

Given the following data: $$2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\Delta G^{\circ}=-6399 \mathrm{kJ}$$ $$\mathrm{C}(s)+\mathrm{o}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta G^{\circ}=-394 \mathrm{kJ}$$ $$\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta G^{\circ}=-237 \mathrm{kJ}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$

Short Answer

Expert verified
The Gibbs free energy change, \(\Delta G^\circ\), for the desired reaction \(6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)\) is +124.5 kJ.

Step by step solution

01

Analyze the given reactions and the desired reaction

The goal is to arrive at the desired reaction using the given reactions. Our given reactions are: 1. 2 C₆H₆(l) + 15 O₂(g) → 12 CO₂(g) + 6 H₂O(l) \(ΔG^\circ = -6399 kJ\) 2. C(s) + O₂(g) → CO₂(g) \(ΔG^\circ = -394 kJ\) 3. H₂(g) + 1/2 O₂(g) → H₂O(l) \(ΔG^\circ = -237 kJ\)
02

Reverse given reaction 1

In order to isolate C₆H₆(l) in the desired reaction, we will reverse reaction 1: 1. 12 CO₂(g) + 6 H₂O(l) → 2 C₆H₆(l) + 15 O₂(g) \(ΔG_r^\circ = +6399 kJ\) Now, the equation looks like this: 6 C(s) + 3 H₂(g) → 2 C₆H₆ (l)
03

Divide reaction 1 by 2

In order to have only one molecule of C₆H₆ on the right side of the equation, we will divide the reversed 1st equation by 2: 1. 6 CO₂(g) + 3 H₂O(l) → C₆H₆(l) + 7.5 O₂(g) \(ΔG_r^\circ = +3199.5 kJ\) Now, the equation looks like this: 6 C(s) + 3 H₂(g) → C₆H₆ (l)
04

Convert CO₂ and H₂O to their constituents

Now, we would like to rewrite CO₂ and H₂O in terms of C and H₂. We will do that using reactions 2 (for CO₂) and 3 (for H₂O), because they provide \(\Delta G^\circ\) for obtaining C and H₂ from CO₂ and H₂O.
05

Multiply given reaction 2 by 6

Since we have 6 moles of CO₂ in the first reaction, we will multiply reaction 2 by 6 to obtain 2. 6 C(s) + 6 O₂(g) → 6 CO₂(g) \(ΔG^\circ = -394 \times 6 = -2364 kJ\)
06

Multiply given reaction 3 by 3

Since we have 3 moles of H₂O in the first reaction, we will multiply reaction 3 by 3 to obtain 3. 3 H₂(g) + 1.5 O₂(g) → 3 H₂O(l) \(ΔG^\circ = -237 \times 3 = -711 kJ\)
07

Combine the reactions

Now adding reactions 2 and 3 to the reversed and divided 1st reaction, we will cancel out CO₂ and H₂O: (+3199.5 kJ): 6 CO₂(g) + 3 H₂O(l) → C₆H₆(l) + 7.5 O₂(g) (+(-2364 kJ)): 6 C(s) + 6 O₂(g) → 6 CO₂(g) (+(-711 kJ)): 3 H₂(g) + 1.5 O₂(g) → 3 H₂O(l) Sum reaction: 6 C(s) + 3 H₂(g) → C₆H₆(l)
08

Calculate the Gibbs free energy change for the desired reaction

With the sum of the reactions being the desired reaction, we can now calculate ΔG⁰ by taking the sum of all energy changes : ΔG⁰ = (+3199.5) + (-2364) + (-711) = +124.5 kJ Thus, the Gibbs free energy change, ΔG⁰, for the desired reaction is +124.5 kJ.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

The melting point for carbon diselenide \(\left(\mathrm{CSe}_{2}\right)\) is \(-46^{\circ} \mathrm{C}\) . At a temperature of \(-75^{\circ} \mathrm{C},\) predict the signs for \(\Delta S_{\mathrm{surr}}\) and $\Delta S_{\mathrm{univ}}\( for the following process: \)\operatorname{CSe}_{2}(l) \rightarrow \operatorname{CSe}_{2}(s)$

For the reaction at 298 K, $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ 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 \(\mathrm{K}\) ? Assuming that \(\Delta H^{p}\) 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?

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 $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ 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. $[\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}$ b. $[\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M$ c. $[\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}$ d. $[\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M$ e. $[\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}$ Based on the calculated DG values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

The following reaction occurs in pure water: $$\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ which is often abbreviated as $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\) . Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 M\) and \(\left[\mathrm{H}^{+}\right]=0.71 M .\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free