/*! 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} Q16.4-36E Is the formation of ozone \(\lef... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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

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

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

Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at \(850\;{\rm{K}}\).

\({\rm{C}}{{\rm{H}}_4}(g) + 4{\rm{C}}{{\rm{l}}_2}(g) \to {\rm{CC}}{{\rm{l}}_4}(g) + 4{\rm{HCl}}(g)\)

What is the equilibrium constant for the reaction at \(850\;{\rm{K}}\)? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?

Consider the decomposition of red mercury(II) oxide under standard state conditions.

\({\bf{2HgO(s, red )}} \to {\bf{2Hg(l) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.