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Under what conditions is \({{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(g)}} \to {\bf{NO(g) + N}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\) spontaneous?

Short Answer

Expert verified

The reaction will be spontaneous at temperatures higher than \(285.17\;{\rm{K}}\).

Step by step solution

01

Determining the spontaneous reaction.

  • The definition of spontaneous reaction in science is a reaction that occurs in a specified set of conditions without interference. A spontaneous reaction is one that occurs in the absence of intervention in a given set of conditions. A spontaneous reaction is completed without the assistance of anyone else.
  • A reaction is spontaneous if the overall entropy, or disorder, increases, according to the Second Law of Thermodynamics.
02

Calculating \({\bf{\Delta H}}\) and \({\bf{\Delta S}}\)

In order to determine under which conditions will the reaction be spontaneous, we must determine its \(\Delta H\)and \(\Delta S\)using the formulas

\(\Delta H = {H_f}({\rm{ products }}) - {H_f}({\rm{ reactants }})\)

and

\(\Delta S = S({\rm{ products }}) - S({\rm{ reactants }})\)

We can look up the values for formation enthalpy change and standard state entropies in the table in Appendix G in the book.

\(\Delta H = {H_f}(NO,g) + {H_f}\left( {N{O_2},g} \right) - {H_f}\left( {{N_2}{O_3},g} \right)\)

\(\Delta H = (90.25 + 33.2 - 83.72)\frac{{kJ}}{{mol}}\)

\(\Delta H = 39.7\frac{{kJ}}{{mol}}\)

\(\Delta S = S(NO,g) + S\left( {N{O_2},g} \right) - S\left( {{N_2}{O_3},g} \right)\)

\(\Delta S = (210.8 + 240.1 - 312.17)\frac{J}{{K \cdot mol}}\)

\(\Delta S = (210.8 + 240.1 - 312.17)\frac{J}{{K \cdot mol}}\)

We have discovered that both the \(\Delta H\) and the \(\Delta S\)are positive. If a reaction is spontaneous, its \(\Delta G\) is negative. If we take a look at the formula for \(\Delta G\) :

\(\Delta G = \Delta H - T\Delta S\)

and have in mind that both \(\Delta H\)and \(\Delta S\)are positive, we can see that the reaction will only be spontaneous at higher temperatures because then will be greater than \(\Delta H\) and \(\Delta G\)will be negative. We can also calculate at which temperature will the reaction become spontaneous:

\(T = \frac{{\Delta H - \Delta G}}{{\Delta S}}\)

\(T = \frac{{39700J_{mol}^{ - 1} - 0.Jmo{l^{ - 1}}}}{{138.73J{K^{ - 1}}mo{l^{ - 1}}}}\)

\(T = 285.17\;{\rm{K}}\)

So the reaction will be spontaneous at temperatures higher than \(285.17\;{\rm{K}}\).

The reaction will be spontaneous at temperatures higher than \(285.17\;{\rm{K}}\).

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Most popular questions from this chapter

Calculate ฮ”Gยฐ using

(a) free energies of formation and

(b) enthalpies of formation and entropies(Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 ยฐC?

\({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{H}}_{\bf{2}}}{\bf{(g) + }}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}}\)

Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain.

(a) \({{\bf{N}}_{\bf{2}}}{\bf{(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}}\)

(b) \({\bf{HCl(g) + N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}} \to {\bf{N}}{{\bf{H}}_{\bf{4}}}{\bf{Cl(s)}}\)

(c) \({\left( {{\bf{N}}{{\bf{H}}_{\bf{4}}}} \right)_{\bf{2}}}{\bf{C}}{{\bf{r}}_{\bf{2}}}{{\bf{O}}_{\bf{7}}}{\bf{(s)}} \to {\bf{C}}{{\bf{r}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s) + 4}}{{\bf{H}}_{\bf{2}}}{\bf{O(g) + }}{{\bf{N}}_{\bf{2}}}{\bf{(g)}}\)

(d) \({\bf{2Fe(s) + 3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{F}}{{\bf{e}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s)}}\)

Calculate the standard entropy change for the following reaction:

\({\bf{Ca(OH}}{{\bf{)}}_{\bf{2}}}{\bf{(\;s)}} \to {\bf{CaO(s) + }}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}}\)

Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.

One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):

\({\text{G}}6{\text{P}} \rightleftharpoons {\text{F}}6{\text{P}}\;\;\;\Delta G_{298}^\circ = 1.7\;{\text{kJ}}\)

(a) Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?

(b) Standard thermodynamic conditions imply the concentrations of G6P and F6P to be \(1M\), however, in a typical cell, they are not even close to these values. Calculate \(\Delta G\)when the concentrations of G6P and F6P are \(120\mu M\) and \(28\mu M\)respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37ยฐC

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