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Is the formation of ozone \(\left( {{{\bf{O}}_{\bf{3}}}{\bf{(g)}}} \right)\) from oxygen \(\left( {{{\bf{O}}_{\bf{2}}}{\bf{(g)}}} \right)\) spontaneous at room temperature under standard state conditions?

Short Answer

Expert verified

The formation of ozone from oxygen is not spontaneous under standard state conditions.

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

Formation of ozone from oxygen.

The equation for the formation of ozone from oxygen:

\(3{{\rm{O}}_2}(g) \to 2{{\rm{O}}_3}(g)\)

In order to determine whether this reaction is spontaneous under standard state conditions, the easiest way would be to calculate the Gibbs energy change\((\Delta G)\). If the \(\Delta G\)is less than zero, it means that the reaction Is spontaneous under standard state conditions. Otherwise, it is not spontaneous under these conditions. To determine the \(\Delta G\)for this reaction, we can use the formula

\(\Delta G = {G_f}({\rm{ products }}) - {G_f}({\rm{ reactants }})\)

\(\Delta G = 2{G_f}\left( {{O_3},g} \right) - 3{G_f}\left( {{O_2},g} \right)\)

\(\Delta G = (2 \cdot 163.2 - 3 \cdot 0)\frac{{kJ}}{{mol}}\)

\(\Delta G = 326.4\frac{{kJ}}{{mol}}\)

Since the \(\Delta G\)is positive, this reaction is not spontaneous under standard state conditions.

Therefore the formation of ozone from oxygen is not spontaneous under standard state conditions.

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Most popular questions from this chapter

Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.

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

(b) Diborane\({\bf{:}}{\rm{ }}{{\bf{B}}_{\bf{2}}}{{\bf{H}}_{\bf{6}}}{\bf{(g)}} \to {\bf{2\;B(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(c) Hydrazine: \({{\bf{N}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{N}}_{\bf{2}}}{\bf{(g) + 2}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(d) Hydrogen peroxide: \({{\bf{H}}_{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(l)}} \to {{\bf{H}}_{\bf{2}}}{\bf{O(g) + }}\frac{{\bf{1}}}{{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

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

Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction.

(a) \({\bf{NaN}}{{\bf{O}}_{\bf{3}}}{\bf{(s)}} \to {\bf{N}}{{\bf{a}}^{\bf{ + }}}{\bf{(aq) + N}}{{\bf{O}}_{\bf{3}}}^{\bf{ - }}{\bf{(aq)}}\)

(b) the freezing of liquid water

(c) \({\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(s)}} \to {\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

(d) \({\bf{CaCO(s)}} \to {\bf{CaO(s) + C}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

Write conversion factors (as ratios) for the number of:

(a) yards in 1 meter

(b) liters in 1 liquid quart

(c) pounds in 1 kilogram

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 37°C

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