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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

The reaction \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\) is highly spontaneous. A classmate calculates the entropy change for this reaction and obtains a large negative value for \(\Delta S^{\circ}\). Did your classmate make a mistake in the calculation? Explain.

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?

Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (i) spontanous at all temperatures; (ii) not spontaneous at any temperature; (iii) spontaneous at low \(T\) but not spontaneous at high \(T ;\) (iv) spontaneous at high T but not spontaneous at low \(T\). $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g) \\ \Delta H^{\circ}=-249 \mathrm{~kJ} ; \Delta S^{\circ}=-278 \mathrm{~J} / \mathrm{K} \\ \text { (b) } \mathrm{N}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NCl}_{3}(g) \\ \Delta H^{\circ}=460 \mathrm{~kJ} ; \Delta S^{\circ}=-275 \mathrm{~J} / \mathrm{K} \\ \text { (c) } \mathrm{N}_{2} \mathrm{~F}_{4}(g) \longrightarrow 2 \mathrm{NF}_{2}(g) \\ \Delta H^{\circ}=85 \mathrm{~kJ} ; \Delta S^{\circ}=198 \mathrm{~J} / \mathrm{K} \end{array} $$

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of \(\mathrm{SO}_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at \(298 \mathrm{~K} ?(\mathbf{b})\) In principle, is this reaction a feasible method of removing \(\mathrm{SO}_{2}\) ? (c) If \(P_{5 \mathrm{O}_{2}}=P_{\mathrm{H}_{2}}\) s and the vapor pressure of water is \(3.33 \mathrm{kPa}\), calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at \(298 \mathrm{~K}\). (d) Would you expect the process to be more or less effective at higher temperatures?

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) \(\mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)+\mathrm{H}_{2}(g)\)

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