/*! 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} Q17E The initial concentrations or pr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Q17E (page 751)

The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction) in which each system will proceed to leach equilibrium.

Short Answer

Expert verified
  1. As a result, the reaction will go in the opposite direction \({Q_C} > {K_c}\)
  2. As a result, the response will move forward\(Q\_\left\{ p \right\} p\)
  3. As a result, the reaction will go in the opposite direction,\({Q_C} > {K_C}\)
  4. As a result, the response will move forward,\(Q\_\left\{ p \right\} p\)
  5. As a result, the response will move forward\(Q\_\left\{ C \right\} c\)
  6. As a result, the response will move forward \(Q\_\left\{ p \right\} p\)

Step by step solution

01

Definition of reaction quotient

  • Under non-equilibrium conditions, the reaction quotient 'Q' is defined as the ratio of initial product concentrations to initial reactant concentrations.
02

Determine the reaction quotient of \(2N{H_3}(g) \rightleftharpoons {N_2}(g) + 3{H_2}(g)\)

(a)

Because of the reaction,\(2N{H_3}(g) \rightleftharpoons {N_2}(g) + 3{H_2}(g)\)

\({Q_c} = \frac{{\left| {{N_2}} \right|{{\left| {{H_2}} \right|}^3}}}{{{{\left| {N{H_3}} \right|}^2}}}\)

Given,

\(\begin{aligned}{}{K_c} = 17;\left( {N{H_3}} \right) = 0.20M,\left( {{N_2}} \right) = 1.00M,\left( {{H_2}} \right) = 1.00M\\{Q_c} = \frac{{(1.00M) \times {{(1.00M)}^3}}}{{{{(0.20M)}^2}}} = 25\end{aligned}\)

As a result, the reaction will go in the opposite direction \({Q_C} > {K_c}\)

03

Determine the reaction quotient of \(2N{H_3}(g)\rightleftharpoons  {N_2}(g) + 3{H_2}(g)1.0\;atm\) 

(b)

Because of the reaction\(2N{H_3}(g)\rightleftharpoons {N_2}(g) + 3{H_2}(g)1.0\;atm\)

\({Q_p} = \frac{{\left( {p{N_2}} \right){{\left( {p{H_2}} \right)}^3}}}{{{{\left( {pN{H_3}} \right)}^2}}}\)

Given,

\(\begin{aligned}{}{{\rm{K}}_{\rm{p}}} = 6.8 \times {10^4};{\rm{pN}}{{\rm{H}}_3} = 3.0\;{\rm{atm}},{\rm{p}}{{\rm{N}}_2} = 2.0\;{\rm{atm}},{\rm{p}}{{\rm{H}}_2} = 1.0\;{\rm{atm}}\\{{\rm{Q}}_{\rm{p}}} = \frac{{(2.0\;{\rm{atm}}){{(1.0\;{\rm{atm}})}^3}}}{{{{(3.0\;{\rm{atm}})}^2}}} = 0.22\end{aligned}\)

As a result, the response will move forward \(Q\_\left\{ p \right\} p\)

04

Determine the reaction quotient of \(2S{O_3}(g)\rightleftharpoons 2S{O_2}(g) + {O_2}(g)\) 

(c)

Because of the reaction,\(2S{O_3}(g)\rightleftharpoons 2S{O_2}(g) + {O_2}(g)\)

\({Q_c} = \frac{{\left. {{{\left| {S{O_2}} \right|}^2}\mid {O_2}} \right)}}{{\left( {{{\left. {S{O_3}} \right|}^2}} \right.}}\)

Given,

\(\begin{aligned}{l}{K_c} = 0.230;\left( {S{O_3}} \right) = 0.00M,\left( {S{O_2}} \right) = 1.00M,\left( {{O_2}} \right) = 1.00M\\{Q_c} = \frac{{{{(1.00M)}^2} \times (1.00M)}}{{{{(0.00M)}^2}}} = \propto \end{aligned}\)

As a result, the reaction will go in the opposite direction, \({Q_C} > {K_C}\)

05

Determine the reaction quotient of \(2S{O_3}(g) \rightleftharpoons 2S{O_2}(g) + {O_2}(g)1.00\;atm\) 

(d)

Because of the reaction

\(2S{O_3}(g) \rightleftharpoons 2S{O_2}(g) + {O_2}(g)1.00\;atm\)

\({Q_p} = \frac{{{{\left( {pS{O_2}} \right)}^2}\left( {p{O_2}} \right)}}{{{{\left( {pS{O_3}} \right)}^2}}}\)

Given,

\(\begin{aligned}{l}{K_p} = 16.5;pS{O_3} = 1.00\;atm,pS{O_2} = 1.00\;atm,p{O_2} = 1.0\;atm\\{Q_p} = \frac{{{{(1.00\;atm)}^2}(1.00\;atm)}}{{{{(1.00\;atm)}^2}}} = 1.0\end{aligned}\)

As a result, the response will move forward, \(Q\_\left\{ p \right\} p\)

06

Determine the reaction quotient of \(2NO(g) + C{l_2}(g) \rightleftharpoons 2NOCl(g)\) 

(e)

Because of the reaction\(2NO(g) + C{l_2}(g) \rightleftharpoons 2NOCl(g)\)

\({Q_c} = \frac{{{{(NOCl)}^2}}}{{\left( {N{O^2}\left( {C{l_2}} \right)} \right.}}\)

Given,

\(\begin{aligned}{}{K_c} = 4.6 \times 1{0^4};(NO) = 1.00M,\left( {C{l_2}} \right) = 1.00M,(NOCl) = 0M\\{Q_c} = \frac{{{{(0M)}^2}}}{{\left( {1.00{M^2}(1.00M)} \right.}} = 0\end{aligned}\)

As a result, the response will move forward \(Q\_\left\{ C \right\} c\).

07

Determine the reaction quotient of \({N_2}(g) + {O_2}(g)  \rightleftharpoons 2NO(g)\)

(f)

Because of the reaction \({N_2}(g) + {O_2}(g) \rightleftharpoons 2NO(g)\)

\({Q_p} = \frac{{\left( {pN{O^2}} \right.}}{{\left( {p{N_2}} \right)\left( {p{O_2}} \right)}}\)

Given,

\(\begin{aligned}{l}{K_p} = 0.050;pNO = 10.0\;atm,p{N_2} = 5\;atm,p{O_2} = 5\;atm\\{Q_p} = \frac{{{{(10.0\;atm)}^2}}}{{(5\;atm)(5\;atm)}} = 4\end{aligned}\)

As a result, the response will move forward \(Q\_\left\{ p \right\} p\)

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 91影视!

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

Which of the systems described in Exercise 13.15 give homogeneous equilibria? Which give heterogeneous equilibria?

(a) \(C{H_4}(g) + C{l_2}\rightleftharpoons C{H_3}CI(g) + HCI(g)\)

(b)\({N_2}(g) + {O_2}(g)\rightleftharpoons 2NO(g)\)

(c)\(2S{O_2}(\;g) + {O_2}(\;g)\rightleftharpoons 2S{O_3}(\;g)\)

(d)\(BaS{O_3}(s)\rightleftharpoons BaO(s) + S{O_2}(g)\)

(e) \({P_4}(g) + 5{O_2}(g)\rightleftharpoons{P_4}{O_{10}}(s)\)

(f)\(B{r_2}(\;g)\rightleftharpoons 2Br(g)\)

(g) \(C{H_4}(g) + 2{O_2}(g)\rightleftharpoons C{O_2}(g) + 2{H_2}O(l)\)

(h) \(CuS{O_4} \times 5{H_2}O(s)\rightleftharpoons CuS{O_4}(s) + 5{H_2}O(g)\)

A sample of ammonium chloride was heated in a closed container. NH4 Cl (s)鈬� NH3 (g) + HCl(g)at equilibrium, the pressure of NH3 (g)was found to be 1.75 atm. What is the value of the equilibrium constant, Kp, for the decomposition at this temperature?

For which of the reactions in Exercise 13.15 does\({K_c}\)(calculated using concentrations) equal\({K_p}\)(calculated using pressures)?

(a) \(C{H_4}(g) + C{l_2} \rightleftharpoons C{H_3}CI(g) + HCI(g)\)

(b) \({N_2}(g) + {O_2}(g)\rightleftharpoons 2NO(g)\)

(c) \(2S{O_2}(\;g) + {O_2}(\;g)\rightleftharpoons 2S{O_3}(\;g)\)

(d) \(BaS{O_3}(s)\rightleftharpoons BaO(s) + S{O_2}(g)\)

(e) \({P_4}(g) + 5{O_2}(g)\rightleftharpoons{P_4}{O_{10}}(s)\)

(f) \(B{r_2}(\;g)\rightleftharpoons 2Br(g)\)

(g) \(C{H_4}(g) + 2{O_2}(g)\rightleftharpoons C{O_2}(g) + 2{H_2}O(l)\)

(h)\(CuS{O_4} \times 5{H_2}O(s)\rightleftharpoons CuS{O_4}(s) + 5{H_2}O(g)\)

Question:The amino acid alanine has two isomers, \(\alpha - alanine\;\)and \(\beta - alanine\;\). When equal masses of these two compounds are dissolved in equal amounts of a solvent, the solution of \(\alpha - alanine\;\)freezes at the lowest temperature. Which form, \(\alpha - alanine\;\)or\(\beta - alanine\;\) has the larger equilibrium constant for ionization \(\left( {HX \rightleftharpoons {H^ + } + {X^ - }} \right)?\)

A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide (\({\rm{S}}{{\rm{O}}_3}\)), from sulfur dioxide (\({\rm{S}}{{\rm{O}}_2}\)), and oxygen (\({{\rm{O}}_2}\)), shown here.

\(2{\text{S}}{{\text{O}}_2}(g) + {{\text{O}}_2}(g) \rightleftharpoons 2{\text{S}}{{\text{O}}_3}(g)\)

At high temperatures, the rate of formation of \({\rm{S}}{{\rm{O}}_3}\)is higher, but the equilibrium amount (concentration or partial pressure) of \({\rm{S}}{{\rm{O}}_3}\) is lower than it would be at lower temperatures.

(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?

(b) Is the reaction endothermic or exothermic?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation
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.