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Chapter 20: Problem 51

Many oxides of nitrogen have positive values for the standard free energy of formation. Using NO as an example, explain why this is the case.

Short Answer

Expert verified
Many oxides of nitrogen, such as NO, have positive values for the standard free energy of formation (螖G掳f) because the enthalpy (螖H掳) and entropy (螖S掳) changes during the formation of these compounds from their constituent elements are both positive. The positive 螖G掳f suggests that the formation of these oxides is non-spontaneous under standard conditions and their stability may be lower compared to their constituent elements, such as N2 and O2. However, the value of 螖G掳f depends on temperature, and the formation of these compounds may become more favorable at higher temperatures.

Step by step solution

01

Introduction to Standard Free Energy of Formation

Standard free energy of formation, denoted by 螖G掳f, is the change in Gibbs free energy when one mole of a substance is formed from its constituent elements in their standard states. Mathematically, it is given by 螖G = 螖H - T螖S, where 螖H is the change in enthalpy, T is the temperature, and 螖S is the change in entropy. A positive 螖G掳f means that the process of forming the compound from its elements is non-spontaneous under standard conditions, indicating that the compound may be less stable compared to its constituent elements. On the other hand, a negative 螖G掳f indicates a spontaneous formation and greater stability of the compound compared to its constituent elements.
02

Formation of NO and its 螖H掳 and 螖S掳 values

Let's consider the formation of nitric oxide (NO) from its constituent elements, nitrogen (N2) and oxygen (O2). The balanced chemical equation for this reaction is: \( \frac{1}{2} N_2 + \frac{1}{2} O_2 \rightarrow NO \) The standard enthalpy change, 螖H掳, for this reaction is positive, since the reaction requires energy to break the strong triple bond in the nitrogen molecule. The standard entropy change, 螖S掳, is also positive for this reaction because one mole of gas is produced while one mole of gas is consumed, which leads to a net increase in the number of gas molecules and thus an increase in entropy.
03

Calculating 螖G掳f for the formation of NO

Now, we can use the equation 螖G = 螖H - T螖S to calculate the standard free energy of formation for NO. Since both 螖H掳 and 螖S掳 are positive, the value of 螖G掳f will depend on the temperature. At lower temperatures, the term "T螖S" will be smaller, making 螖G掳f more likely to be positive. At higher temperatures, the term "T螖S" will increase, and the value of 螖G掳f may become negative, indicating that the formation of NO becomes more favorable.
04

Explanation for the positive 螖G掳f of NO

In summary, many oxides of nitrogen, such as NO, have positive values for the standard free energy of formation because the enthalpy and entropy changes during the formation of these compounds from their constituent elements are both positive. The positive 螖G掳f suggests that the formation of these oxides is non-spontaneous under standard conditions, and their stability may be lower compared to their constituent elements such as N2 and O2. However, it's important to note that the value of 螖G掳f depends on temperature, and the formation of these compounds may become more favorable at higher temperatures.

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Most popular questions from this chapter

EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt \(\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}\), are also used to treat heavy metal poisoning. The equilibrium constant for the following reaction is \(1.0 \times 10^{23}\) : Calculate \(\left[\mathrm{Pb}^{2+}\right]\) at equilibrium in a solution originally \(0.0010 \mathrm{M}\) in \(\mathrm{Pb}^{2+}, 0.050 \mathrm{M}\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\), and buffered at \(\mathrm{pH}=6.00\).

Why are the tin(IV) halides more volatile than the tin(II) halides?

Use bond energies (Table 8.4) to show that the preferred products for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{3}\) are \(\mathrm{NO}_{2}\) and \(\mathrm{NO}\) rather than \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}\). (The \(\mathrm{N}-\mathrm{O}\) single bond energy is \(201 \mathrm{~kJ} / \mathrm{mol} .\) ) Hint: Consider the reaction kinetics.

One pathway for the destruction of ozone in the upper atmosphere is $$\begin{array}{l}\mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \text { Slow } \\\\\mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) \quad \text { Fast } \\ \text { Overall reaction: } \mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\end{array}$$ a. Which species is a catalyst? b. Which species is an intermediate? c. The activation energy \(E_{\mathrm{a}}\) for the uncatalyzed reaction $$\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)$$ is \(14.0 \mathrm{~kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed by the presence of \(\mathrm{NO}\) is \(11.9 \mathrm{~kJ} .\) What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C}\) ? Assume that the frequency factor \(A\) is the same for each reaction. d. One of the concerns about the use of Freons is that they will migrate to the upper atmosphere, where chlorine atoms can be generated by the reaction $$\mathrm{CCl}_{2} \mathrm{~F}_{2} \stackrel{\mathrm{hr}}{\longrightarrow} \mathrm{CF}_{2} \mathrm{Cl}+\mathrm{Cl}$$ Freon- 12 Chlorine atoms also can act as a catalyst for the destruction of ozone. The first step of a proposed mechanism for chlorinecatalyzed ozone destruction is $$\mathrm{Cl}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)$$ Slow Assuming a two-step mechanism, propose the second step in the mechanism and give the overall balanced equation. e. The activation energy for Cl-catalyzed destruction of ozone is \(2.1 \mathrm{~kJ} / \mathrm{mol}\). Estimate the efficiency with which \(\mathrm{Cl}\) atoms destroy ozone as compared with NO molecules at \(25^{\circ} \mathrm{C}\). Assume that the frequency factor \(A\) is the same for each catalyzed reaction and assume similar rate laws for each catalyzed reaction.

The most significant source of natural radiation is radon- \(222 .\) \({ }^{222} \mathrm{Rn}\), a decay product of \({ }^{2.38} \mathrm{U}\), is continuously generated in the earth's crust, allowing gaseous \(\mathrm{Rn}\) to seep into the basements of buildings. Because \({ }^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of \(3.82\) days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \({ }^{238} \mathrm{U}\) decays to \({ }^{222} \mathrm{Rn}\) ? What nucleus is produced when \({ }^{222} \mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \({ }^{222} \mathrm{Rn}\) ? -
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