/*! 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 69 Consider the following reaction ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following reaction between oxides of nitrogen: $$ \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{NO}(g) $$ (a) Use data in Appendix \(C\) to predict how \(\Delta G\) for the reaction varies with increasing temperature. (b) Calculate \(\Delta G\) at \(800 \mathrm{~K}\), assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with temperature. Under standard conditions is the reaction spontaneous at \(800 \mathrm{~K} ?\) (c) Calculate \(\Delta G\) at \(1000 \mathrm{~K}\). Is the reaction spontaneous under standard conditions at this temperature?

Short Answer

Expert verified
(a) The Gibbs Free Energy equation is given by \(\Delta G = \Delta H - T \Delta S\). (b) At 800 K, we have \(\Delta H = 55.8 \: kJ\), \(\Delta S = 173.5 \: J/K\), and \(\Delta G = -84,000 \: J/mol\). The reaction is spontaneous at 800 K, as \(\Delta G\) is negative. (c) At 1000 K, \(\Delta G = -117,700 \: J/mol\), which means the reaction is also spontaneous under standard conditions at 1000 K.

Step by step solution

01

a) Knowing the Gibbs Free Energy Equation:

We need to know the Gibbs Free Energy equation which is as follows: \[ \Delta G = \Delta H - T \Delta S \] where ∆G is the change in Gibbs free energy, ∆H is the change in enthalpy, T is the temperature in Kelvin, and ∆S is the change in entropy.
02

b) Calculating ∆G at 800 K:

We are given that ∆H° and ∆S° do not change with temperature. Using the data given in Appendix C, we have: ∆H°(NO2) = 33.1 kJ/mol, ∆S°(NO2) = 240.0 J/mol·K ∆H°(N2O) = 82.0 kJ/mol, ∆S°(N2O) = 218.6 J/mol·K ∆H°(NO) = 90.3 kJ/mol, ∆S°(NO) = 210.7 J/mol·K Now, we have to calculate the ∆H and ∆S of the reaction by using stoichiometric coefficients. ∆H = [3∆H°(NO)] - [∆H°(NO2) + ∆H°(N2O)] = (3*90.3) - (33.1+82.0) = 55.8 kJ ∆S = [3∆S°(NO)] - [∆S°(NO2) + ∆S°(N2O)] = (3*210.7) - (240.0+218.6) = 173.5 J/K Now we can calculate ∆G at 800 K: ∆G = ∆H - T∆S = 55.8*10^3 J/mol - (800 K)(173.5 J/mol·K) = -84,000 J/mol As ∆G is negative, the reaction is spontaneous at 800 K.
03

c) Calculating ∆G at 1000 K:

We already have ∆H and ∆S for the reaction; we only need to change the temperature in the equation for Gibbs Free Energy. ∆G = ∆H - T∆S = 55.8*10^3 J/mol - (1000 K)(173.5 J/mol·K) = -117,700 J/mol As ∆G is negative at 1000 K, the reaction is spontaneous under standard conditions at this temperature as well.

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Key Concepts

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

Spontaneous Reaction
A reaction is considered spontaneous if it occurs on its own without any external input of energy once it has begun. The spontaneity of a reaction is often determined by the change in Gibbs Free Energy (\(\Delta G\)). When \(\Delta G\) is negative, the reaction proceeds spontaneously. This negativity indicates that the reaction releases energy, typically resulting in a more stable, lower energy state of the system.
  • Negative \(\Delta G\): Spontaneous reaction (energy is released)
  • Positive \(\Delta G\): Non-spontaneous reaction (energy is required)
  • Zero \(\Delta G\): Reaction at equilibrium (no net change occurs)
The calculation of \(\Delta G\) is essential in predicting reaction spontaneity. It's important to know that temperature can affect \(\Delta G\), making some reactions spontaneous only under certain conditions. As seen with the reaction at 800 K, the calculated \(\Delta G\) was negative, heralding a spontaneous process.
Enthalpy Change
Enthalpy change (\(\Delta H\)) is the amount of heat released or absorbed during a chemical reaction at constant pressure. It's a crucial factor in the determination of reaction spontaneity and can be either positive or negative.
  • Exothermic reaction (\(\Delta H < 0\)): Releases heat, contributing to a potentially negative \(\Delta G\).
  • Endothermic reaction (\(\Delta H > 0\)): Absorbs heat, which may require careful consideration regarding \(\Delta G\).
For the reaction provided, calculating \(\Delta H\) involved considering the enthalpy changes of reactants and products. Using stoichiometric coefficients, the reaction enthalpy at different temperatures was determined, aiding in the prediction of its spontaneity at a higher temperature like 1000 K.
Entropy Change
Entropy change (\(\Delta S\)) reflects the degree of disorder or randomness in a system. A positive \(\Delta S\) indicates an increase in entropy or disorder, often favoring spontaneity in a reaction.
  • Positive \(\Delta S\): Greater disorder, often leads to a negative \(\Delta G\).
  • Negative \(\Delta S\): Lesser disorder, could increase \(\Delta G\).
The entropy change in our reaction was calculated using the sum of the entropy changes of the products minus the sum of the entropies of the reactants. The result was a positive entropy change, suggesting that the system becomes more disordered during the reaction. This positive influence on \(\Delta G\) at different temperatures, like at 800 K and 1000 K, demonstrates its critical role in making reactions more spontaneous.

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Most popular questions from this chapter

Using data from Appendix \(\mathrm{C}\), calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions. (a) \(2 \mathrm{Zn}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)\) (b) \(2 \mathrm{NaBr}(s) \longrightarrow 2 \mathrm{Na}(g)+\mathrm{Br}_{2}(g)\) (c) \(\mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)+\mathrm{H}_{2}(g)\)

The normal boiling point of the element mercury (Hg) is \(356.7{ }^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=59.11 \mathrm{~kJ} / \mathrm{mol} .\) (a) When Hg boils at its nor- mal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(2.00 \mathrm{~mol}\) of \(\mathrm{Hg}\) is vaporized at \(356.7^{\circ} \mathrm{C}\).

Which of the following processes are spontaneous: (a) the evaporation of water at \(\$ T P\) to form water vapor of 101.3 kPa pressure; (b) separation of a mixture of water and oil into two separate phases; (c) the souring of milk; (d) the neutralization of hydrochloric acid with sodium hydroxide at \(\mathrm{STP} ;(\mathbf{e})\) the formation of ice from water at \(20^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa} ?\)

The following processes were all discussed in Chapter 18 , "Chemistry of the Environment." Estimate whether the entropy of the system increases or decreases during each process: (a) photodissociation of \(\mathrm{O}_{2}(g),(\mathbf{b})\) formation of ozone from oxygen molecules and oxygen atoms, (c) diffusion of CFCs into the stratosphere, (d) desalination of water by reverse osmosis.

The normal boiling point of \(n\) -octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is \(125^{\circ} \mathrm{C}\). (a) Is the condensation of gaseous \(n\) -octane to liquid \(n\) -octane an endothermic or exothermic process? (b) In what temperature range is the boiling of \(n\) -octane a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid \(n\) -octane and gaseous \(n\) -octane are in equilibrium? Explain.

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