/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Consider the reaction \(2 \mathr... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (a) Using data from Appendix \(\mathrm{C},\) calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of all gases are \(33.4 \mathrm{kPa}\).

Short Answer

Expert verified
(a) The standard Gibbs free energy change (∆G°) at 298 K is -171.2 kJ/mol. (b) The non-standard Gibbs free energy change (∆G) at 298 K with the given partial pressures is -142627.28 J/mol, which is approximately -142.63 kJ/mol.

Step by step solution

01

Calculate standard Gibbs free energy change

Using the ∆G° value for each species provided in Appendix C, we can calculate the standard Gibbs free energy change for the reaction. We have the reaction: 2 NO(g) + O₂(g) ⟶ 2 NO₂(g) Using the formula mentioned above, we get: ∆G° = Σ(ν_product * ∆G°_product) - Σ(ν_reactant * ∆G°_reactant) ∆G° = (2 * ∆G°(NO₂) - (2 * ∆G°(NO) + ∆G°(O₂)) Plug in the values for each species: ∆G° = (2 * 51.3 kJ/mol) - (2 * 86.6 kJ/mol + 0 kJ/mol) Calculate the Gibbs free energy change: ∆G° = -171.2 kJ/mol
02

Calculate reaction quotient

We are given that the partial pressures of all gases are 33.4 kPa. Calculate the reaction quotient Q using the partial pressures: Q = [NO₂]² / ([NO]² * [O₂]) Since the partial pressures are the same for each gas (33.4 kPa), we can write: Q = (33.4)² / ((33.4)² * 33.4) Q = 1 / 33.4
03

Calculate non-standard Gibbs Free Energy Change

Now, we can calculate the non-standard Gibbs free energy change (∆G) using the Van't Hoff equation: ∆G = ∆G° + RT ln(Q) Plug in the values: ∆G = -171.2 kJ/mol + (8.314 J/mol*K * 298 K * ln(1/33.4)) Don't forget to convert kJ to J: ∆G = -171200 J/mol + (8.314 J/mol*K * 298 K * ln(1/33.4)) Calculate the value to obtain the final answer: ∆G = -142627.28 J/mol In conclusion, (a) The standard Gibbs free energy change (∆G°) at 298 K is -171.2 kJ/mol. (b) The non-standard Gibbs free energy change (∆G) at 298 K with the given partial pressures is -142627.28 J/mol, which is approximately -142.63 kJ/mol.

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Key Concepts

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

Chemical Thermodynamics
Chemical thermodynamics helps us understand how and why chemical reactions occur. At its core, it's about energy - tracking energy changes in reactions. Gibbs Free Energy (denoted as \(\Delta G\)) is a key part of this.

Gibbs Free Energy combines enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)) to predict whether a reaction is spontaneous. It's defined by the equation:
  • \[ \Delta G = \Delta H - T \Delta S \]
This equation tells us about the feasibility of reactions:
  • If \(\Delta G\) is negative, the reaction is spontaneous and can proceed without added energy.
  • If \(\Delta G\) is positive, the reaction isn't spontaneous and needs energy to proceed.
Calculating \(\Delta G\) involves standard free energy changes (\(\Delta G^\circ\)), which use values at standard conditions (25°C, 1 atm pressure). This helps us predict reaction behavior under these conditions.
Reaction Quotient
The reaction quotient \(Q\) is a snapshot of a reaction's position at any given moment. Calculating \(Q\) involves the concentrations or partial pressures of reactants and products at a specific time.

For the reaction \(2 \text{NO}(g) + \text{O}_2(g) \longrightarrow 2 \text{NO}_2(g)\), the reaction quotient \(Q\) is calculated by:
  • \[ Q = \frac{[\text{NO}_2]^2}{[\text{NO}]^2 [\text{O}_2]} \]
When \(Q\) equals the equilibrium constant \(K\), the reaction is at equilibrium. If \(Q eq K\), the reaction will shift to reach equilibrium:
  • If \(Q
  • If \(Q>K\), the reaction favors the reverse direction (forming reactants).
In non-standard conditions, using \(Q\) helps calculate \(\Delta G\), which guides our predictions about the reaction's direction and spontaneity.
Non-Standard Conditions
Reactions often occur under non-standard conditions, with varying temperatures, pressures, or concentrations. These require adjustments to the standard Gibbs Free Energy, \(\Delta G^\circ\).

Non-standard Gibbs Free Energy (\(\Delta G\)) can be calculated using the expression:
  • \[ \Delta G = \Delta G^\circ + RT \ln(Q) \]
Here, \(R\) is the universal gas constant, \(T\) is temperature in Kelvin, and \(Q\) is the reaction quotient.

This equation adjusts \(\Delta G^\circ\) to account for actual conditions. For our example under non-standard conditions, with specific partial pressures, \(\Delta G\) changes and tells us about the reaction's spontaneity in real life scenarios.
  • A negative \(\Delta G\) indicates spontaneous reactions even at non-standard conditions.
  • A positive \(\Delta G\) suggests the need for additional energy.
Understanding these adjustments allows chemists to predict and control chemical reactions effectively, even when conditions stray from the standard.

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Most popular questions from this chapter

Carbon disulfide \(\left(C S_{2}\right)\) is a toxic, highly flammable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(I)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) \begin{tabular}{lcc} \hline & \(\Delta H_{i}(\mathrm{k} / \mathrm{mol})\) & \(\Delta G_{i}^{\prime}(\mathrm{kJ} / \mathrm{mol})\) \\ \hline\(C S_{2}(l)\) & 89.7 & 65.3 \\ \(C S_{2}(g)\) & 117.4 & 67.2 \\ \hline \end{tabular} (a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the \(\mathrm{C}-\mathrm{S}\) bonds? \((\mathbf{b})\) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) burns in \(\mathrm{O}_{2}\) with a blue flame, forming \(\mathrm{CO}_{2}(g)\) and \(\mathrm{SO}_{2}(g)\). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix \(C,\) calculate \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the reaction in part \((c) .\) Is the reaction exothermic? Is it spontaneous at \(298 \mathrm{~K} ?\) (e) Use the data in the table to calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\) for the vaporization of \(\mathrm{CS}_{2}(I) .\) Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(l)\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(101.3 \mathrm{kPa}\) ?

For the isothermal expansion of a gas into a vacuum, \(\Delta E=0, q=0,\) and \(w=0 .\) (a) Is this a spontaneous process? (b) Explain why no work is done by the system during this process. \((\mathbf{c})\) What is the "driving force" for the expansion of the gas: enthalpy or entropy?

Does the entropy of the system increase, decrease, or stay the same when (a) a solid melts, (b) a gas liquefies, \((\mathbf{c})\) a solid sublimes?

Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about \(88 \mathrm{~J} / \mathrm{mol}-\mathrm{K} .(\) a) Estimate the normal boiling point of bromine, \(\mathrm{Br}_{2}\), by determining \(\Delta H_{\text {vap }}^{\circ}\) for \(\mathrm{Br}_{2}\) using data from Appendix \(C\). Assume that \(\Delta H_{\text {vap }}^{\circ}\) remains constant with temperature and that Trouton's rule holds. (b) Look up the normal boiling point of \(\mathrm{Br}_{2}\) in a chemistry handbook or at the WebElements website (www.webelements.com) and compare it to your calculation. What are the possible sources of error, or incorrect assumptions, in the calculation?

(a) Using data in Appendix \(C\), estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(s)\) to \(\mathrm{I}_{2}(g)\) is zero. (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experimental melting and boiling points of \(I_{2}\). (c) Which of the values in part (b) is closer to the value you obtained in part (a)?

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