Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Chapter 14, Problem 14.87QE

(a)

Interpretation Introduction

Interpretation:

The concentration of all species after the attainment of equilibrium in the following reaction has to be determined.

  2SO3(g)2SO2(g)+O2(g)

Concept Introduction:

The condition of equilibrium is a state of balance of processes that runs in opposite directions. At equilibrium, the formation of product from reactant balances the formation of reactant from product. Also, the change in concentration of reaction and product seems to be negligible at equilibrium state.

The general equilibrium reaction is as follows:

  aA+bBcC+dD

Here,

A and B are the reactants.

C and D are products.

a and b are the stoichiometric coefficients of reactants.

c and d are the stoichiometric coefficients of products.

(a)

Expert Solution
Check Mark

Answer to Problem 14.87QE

The concentration of O2 is 0.0020 M, that of SO2 is 0.0040 M and that of SO3 is 0.001 M.

Explanation of Solution

The given reaction occurs as follows:

  2SO3(g)2SO2(g)+O2(g)

The concentration of SO3 can be calculated as follows:

  [SO3]=0.010 mol2 L=0.005 M

The ICE table for the above reaction is as follows:

Equation2SO32SO2+O2Initial(M)0.00500Change(M)x+2x+xEquilibrium(M)0.005x2xx

The concentration of O2 at equilibrium can be calculated as follows:

  [O2]=0.0040 mol2 L=0.0020 M

The concentration of O2 at equilibrium is 0.0020 M. Therefore the value of x is 0.0020 M.

The equilibrium concentration of SO2 can be calculated as follows:

  [SO2]=2(0.0020 M)=0.0040 M

Therefore the equilibrium concentration of SO2 is 0.0040 M.

The equilibrium concentration of SO3 can be calculated as follows:

  [SO2]=(0.005 M2(0.0020 M))=0.001 M

Therefore the equilibrium concentration of SO3 is 0.001 M.

(b)

Interpretation Introduction

Interpretation:

The value of Kc for the following reaction has to be determined.

  2SO3(g)2SO2(g)+O2(g)

Concept Introduction:

The general equilibrium reaction is as follows:

  aA+bBcC+dD

Here,

A and B are the reactants.

C and D are products.

a and b are the stoichiometric coefficients of reactants.

c and d are the stoichiometric coefficients of products.

The expression of the equilibrium constant for the above reaction is as follows:

  Kc=[C]c[D]d[A]a[B]b

Here,

Kc is the equilibrium constant.

[C] is the concentration of C.

[D] is the concentration of D.

[A] is the concentration of A.

[B] is the concentration of B.

(b)

Expert Solution
Check Mark

Answer to Problem 14.87QE

The value of Kc for the given reaction is 0.032.

Explanation of Solution

The given reaction occurs as follows:

  2SO3(g)2SO2(g)+O2(g)

The expression of Kc for the above reaction is as follows:

  Kc=[SO2]2[O2][SO3]2        (1)

Substitute 0.0040 M for [SO2], 0.0020 M for [O2] and 0.001 M for [SO3] in equation (1).

  Kc=(0.0040)2(0.0020)(0.001)2=0.032

(c)

Interpretation Introduction

Interpretation:

The value of Kp for the following reaction has to be determined.

  2SO3(g)2SO2(g)+O2(g)

Concept Introduction:

The relation of equilibrium constant Kp and Kc is as follows:

  Kp=Kc(RT)Δn        (2)

Here, Δn is change in number of moles of gas.

The formula to calculate the value of Δn is as follows:

  Δn=(total number of moles ofgas on the product side)(total number of moles ofgas on the reactant side)        (3)

The value of Δn can be zero, negative and positive.

(c)

Expert Solution
Check Mark

Answer to Problem 14.87QE

The value of Kp for the given reaction is 2.9.

Explanation of Solution

The given reaction occurs as follows:

  2SO3(g)2SO2(g)+O2(g)

Substitute 3 for the total number of moles of gas on product side and 2 for the total number of moles of gas on reactant side in equation (3).

  Δn=32=1

The formula to convert degree Celsius to Kelvin is as follows:

  T(K)=T(°C)+273.15 K        (4)

Substitute 832 °C for T(°C) in equation (4) to calculate T.

  T(K)=832 °C+273.15 K=1105.15 K

Substitute 1 for Δn, 0.032 for Kc, 1105.15 K for T and 0.08206 Latm/molK for R in equation (2).

  Kp=(0.032)((0.08206)(1105.15))1=2.9

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Chapter 14 Solutions

Chemistry: Principles and Practice

Ch. 14 - Explain why terms for pure liquids and solids do...Ch. 14 - Temperature influences solubility. Does...Ch. 14 - Prob. 14.13QECh. 14 - Prob. 14.14QECh. 14 - Prob. 14.15QECh. 14 - Prob. 14.16QECh. 14 - Prob. 14.17QECh. 14 - Prob. 14.18QECh. 14 - At 2000 K, experiments show that the equilibrium...Ch. 14 - At 500 K, the equilibrium constant is 155 for...Ch. 14 - At 77 C, Kp is 1.7 104 for the formation of...Ch. 14 - Consider the following equilibria involving SO2(g)...Ch. 14 - Kc at 137 C is 4.42 for NO(g) + 12 Br2(g) NOBr(g)...Ch. 14 - Prob. 14.24QECh. 14 - Prob. 14.25QECh. 14 - Prob. 14.26QECh. 14 - Prob. 14.27QECh. 14 - Prob. 14.28QECh. 14 - Prob. 14.29QECh. 14 - Prob. 14.30QECh. 14 - Prob. 14.31QECh. 14 - Prob. 14.32QECh. 14 - Prob. 14.33QECh. 14 - Prob. 14.34QECh. 14 - Prob. 14.35QECh. 14 - Consider the system...Ch. 14 - Prob. 14.37QECh. 14 - Prob. 14.38QECh. 14 - Prob. 14.39QECh. 14 - Prob. 14.40QECh. 14 - Prob. 14.41QECh. 14 - Prob. 14.42QECh. 14 - Prob. 14.43QECh. 14 - Prob. 14.44QECh. 14 - Prob. 14.45QECh. 14 - Prob. 14.46QECh. 14 - Prob. 14.47QECh. 14 - Prob. 14.48QECh. 14 - Prob. 14.49QECh. 14 - Prob. 14.50QECh. 14 - Prob. 14.51QECh. 14 - Consider 0.200 mol phosphorus pentachloride sealed...Ch. 14 - Prob. 14.53QECh. 14 - Prob. 14.54QECh. 14 - Prob. 14.55QECh. 14 - Prob. 14.56QECh. 14 - Prob. 14.57QECh. 14 - Prob. 14.58QECh. 14 - Prob. 14.59QECh. 14 - Prob. 14.60QECh. 14 - Prob. 14.61QECh. 14 - Write the expression for the equilibrium constant...Ch. 14 - Prob. 14.63QECh. 14 - Prob. 14.64QECh. 14 - Write the expression for the solubility product...Ch. 14 - Prob. 14.66QECh. 14 - Prob. 14.67QECh. 14 - The solubility of silver iodate, AgIO3, is 1.8 ...Ch. 14 - Prob. 14.69QECh. 14 - Prob. 14.70QECh. 14 - Prob. 14.71QECh. 14 - Prob. 14.72QECh. 14 - Even though barium is toxic, a suspension of...Ch. 14 - Lead poisoning has been a hazard for centuries....Ch. 14 - Calculate the solubility of barium sulfate (Ksp =...Ch. 14 - Calculate the solubility of copper(II) iodate,...Ch. 14 - Calculate the solubility of lead fluoride, PbF2...Ch. 14 - Calculate the solubility of zinc carbonate, ZnCO3...Ch. 14 - Prob. 14.79QECh. 14 - Prob. 14.80QECh. 14 - Use the solubility product constant from Appendix...Ch. 14 - Prob. 14.82QECh. 14 - Some barium chloride is added to a solution that...Ch. 14 - Prob. 14.84QECh. 14 - Prob. 14.85QECh. 14 - Prob. 14.86QECh. 14 - Prob. 14.87QECh. 14 - Prob. 14.88QECh. 14 - Prob. 14.89QECh. 14 - Prob. 14.90QECh. 14 - Prob. 14.91QECh. 14 - At 3000 K, carbon dioxide dissociates CO2(g) ...Ch. 14 - Prob. 14.94QECh. 14 - Nitrogen, hydrogen, and ammonia are in equilibrium...Ch. 14 - The concentration of barium in a saturated...Ch. 14 - According to the Resource Conservation and...Ch. 14 - Prob. 14.98QE
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