Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 17, Problem 17.87P

(a)

Interpretation Introduction

Interpretation:

The oxidation of SO2 in the manufacture of sulfuric acid determines that Kc = 1.7×188 at 600.K:

  2SO2(g)+ O2(g)2SO3(g)

At equilibrium, PSO3= 300.atm and PO2= 100.atmPSO2 has to be calculated.

Concept Introduction:

Equilibrium constant using partial pressure:

The equilibrium constant of partial pressure can be defined as the ratio of products and reactants concentration at equilibrium in terms of partial pressure.

For a reaction,

  aA(g)+ bB(g)cC(g)+ dD(g)

The expression of Kp can be given as

  Kp = (PC)c(PD)d(PA)a(PB)b

(a)

Expert Solution
Check Mark

Explanation of Solution

Given, Kc = 1.7×188 at 600.K.

  Kp= Kc(RT)Δnwhere, Δn = moles of gaseous products - moles of gaseous reactants Δn = 2 - 3 = -1

  Kp= (1.7×108)[(0.0821 L.atm/mol.K)(600.K)]-1Kp = 3.451×106

At equilibrium,

  KpPSO32PSO22PO2Kp(300)2PSO22Kp= 3.451×106PSO2= 0.016149 atm

PSO2 is calculated as 0.016149 atm.

(b)

Interpretation Introduction

Interpretation:

An engineer places a mixture of 0.0040 mol of SO2(g) and 0.0028 mol of O2(g) in a 1.0-L container and raises the temperature to 1000 K.  At equilibrium, 0.0020 mol of SO3(g) is present.  For this reaction, at 1000 K, Kc and PSO2 has to be calculated.

Concept Introduction:

Equilibrium constant:

The relationship between the concentration of products and concentration of reactants in a chemical reaction at equilibrium is said to be equilibrium constant.  It is denoted by K.

For a reaction,

  xX + yY  zZ

The expression of K can be given as

  Kc = [Z]z[X]x[Y]ywhere, [X] = equilibrium concentration of X [Y] = equilibrium concentration of Y [Z] = equilibrium concentration of Z

Pressure can be calculated using ideal gas equation.

  PV =nRTP    =nRTVwhere, P = Pressure V = Volume R = universal gas constant T = temperature n = number of moles

(b)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is

  2SO2(g)+ O2(g)2SO3(g)

Initial:

Concentration of 2SO2(g) is 0.0040.

Concentration of O2(g) is 0.0028.

Concentration of 2SO3(g) is zero.

Change:

Concentration of 2SO2(g) is -2x.

Concentration of O2(g) is -x.

Concentration of 2SO3(g) is +2x.

At equilibrium:

Concentration of 2SO2(g) is 0.0040-2x.

Concentration of O2(g) is 0.0028-x.

Concentration of 2SO3(g) is 2x.

Given, at equilibrium, 0.0020 mol of SO3(g).

  2x = 0.0020x   = 0.0010

By substituting the x value, concentrations of reactants and products at equilibrium are calculated.

Concentration of SO2:

  [SO2] = 0.0040-2x = 0.0040-2(0.0010) = 0.0020 M

Concentration of O2:

  [O2] = 0.0028-x = 0.0028-0.0010 = 0.0018 M

Concentration of SO3:

  [SO3] = 2x = 2(0.0010) = 0.0020 M

By using the obtained concentration values at equilibrium, Kc is calculated.

  Kc = [SO3]2[SO2]2[O2]Kc = [0.0020]2[0.0020]2[0.0018]Kc = 555.5556

At 1000.K, PSO2 can be calculated by ideal gas equation.

  PSO2 = nRTVPSO2 = (0.0020 mol)(0.0821 L.atm/mol.K)(1000.K)(1.0 L)PSO2 = 0.1642 atm

At 1000 K, Kc and PSO2 are

  Kc  = 555.5556PSO2= 0.1642 atm

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Chapter 17 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 17.5 - Prob. 17.6AFPCh. 17.5 - Prob. 17.6BFPCh. 17.5 - Prob. 17.7AFPCh. 17.5 - Prob. 17.7BFPCh. 17.5 - Prob. 17.8AFPCh. 17.5 - Prob. 17.8BFPCh. 17.5 - Prob. 17.9AFPCh. 17.5 - Prob. 17.9BFPCh. 17.5 - Prob. 17.10AFPCh. 17.5 - Prob. 17.10BFPCh. 17.5 - An inorganic chemist studying the reactions of...Ch. 17.5 - A chemist studying the production of nitrogen...Ch. 17.6 - In a study of glass etching, a chemist examines...Ch. 17.6 - Prob. 17.12BFPCh. 17.6 - Prob. 17.13AFPCh. 17.6 - Prob. 17.13BFPCh. 17.6 - Prob. 17.14AFPCh. 17.6 - Should T be increased or decreased to yield more...Ch. 17.6 - Prob. 17.15AFPCh. 17.6 - Prob. 17.15BFPCh. 17.6 - Many metabolites are products in branched...Ch. 17 - Prob. 17.1PCh. 17 - When a chemical company employs a new reaction to...Ch. 17 - If there is no change in concentrations, why is...Ch. 17 - Prob. 17.4PCh. 17 - Prob. 17.5PCh. 17 - Prob. 17.6PCh. 17 - Prob. 17.7PCh. 17 - Prob. 17.8PCh. 17 - Prob. 17.9PCh. 17 - Does Q for the formation of 1 mol of NO from its...Ch. 17 - Does Q for the formation of 1 mol of NH3 from H2...Ch. 17 - Balance each reaction and write its reaction...Ch. 17 - Prob. 17.13PCh. 17 - Prob. 17.14PCh. 17 - Prob. 17.15PCh. 17 - At a particular temperature, Kc = 1.6×10−2...Ch. 17 - Prob. 17.17PCh. 17 - Balance each of the following examples of...Ch. 17 - Balance each of the following examples of...Ch. 17 - Balance each of the following examples of...Ch. 17 - Balance each of the following examples of...Ch. 17 - Prob. 17.22PCh. 17 - Prob. 17.23PCh. 17 - Prob. 17.24PCh. 17 - When are Kc and Kp equal, and when are they not? Ch. 17 - A certain reaction at equilibrium has more moles...Ch. 17 - Prob. 17.27PCh. 17 - Determine Δngas for each of the following...Ch. 17 - Prob. 17.29PCh. 17 - Prob. 17.30PCh. 17 - Prob. 17.31PCh. 17 - Prob. 17.32PCh. 17 - Prob. 17.33PCh. 17 - The following molecular scenes depict the aqueous...Ch. 17 - At 425°C, Kp = 4.18 × 10−9 for the...Ch. 17 - At 100°C, Kp = 60.6 for the reaction 2NOBr(g) ⇌...Ch. 17 - The water-gas shift reaction plays a central role...Ch. 17 - In the 1980s, CFC-11 was one of the most heavily...Ch. 17 - For a problem involving the catalyzed reaction of...Ch. 17 - What is the basis of the approximation that avoids...Ch. 17 - Prob. 17.41PCh. 17 - Gaseous ammonia was introduced into a sealed...Ch. 17 - Prob. 17.43PCh. 17 - Prob. 17.44PCh. 17 - Prob. 17.45PCh. 17 - Prob. 17.46PCh. 17 - Prob. 17.47PCh. 17 - Prob. 17.48PCh. 17 - Prob. 17.49PCh. 17 - Nitrogen dioxide decomposes according to the...Ch. 17 - Hydrogen iodide decomposes according to the...Ch. 17 - Compound A decomposes according to the...Ch. 17 - In an analysis of interhalogen reactivity, 0.500...Ch. 17 - A toxicologist studying mustard gas, S(CH2CH2Cl)2,...Ch. 17 - Prob. 17.55PCh. 17 - A key step in the extraction of iron from its ore...Ch. 17 - What does “disturbance” mean in Le Châtelier’s...Ch. 17 - Prob. 17.58PCh. 17 - Prob. 17.59PCh. 17 - Prob. 17.60PCh. 17 - Le Châtelier’s principle is related ultimately to...Ch. 17 - An equilibrium mixture of two solids and a gas, in...Ch. 17 - Consider this equilibrium system: CO(g) + Fe3O4(s)...Ch. 17 - Sodium bicarbonate undergoes thermal decomposition...Ch. 17 - Prob. 17.65PCh. 17 - Prob. 17.66PCh. 17 - Predict the effect of decreasing the container...Ch. 17 - Prob. 17.68PCh. 17 - Prob. 17.69PCh. 17 - Prob. 17.70PCh. 17 - Prob. 17.71PCh. 17 - Prob. 17.72PCh. 17 - Prob. 17.73PCh. 17 - The formation of methanol is important to the...Ch. 17 - Prob. 17.75PCh. 17 - The oxidation of SO2 is the key step in H2SO4...Ch. 17 - A mixture of 3.00 volumes of H2 and 1.00 volume of...Ch. 17 - You are a member of a research team of chemists...Ch. 17 - For the following equilibrium system, which of the...Ch. 17 - Prob. 17.80PCh. 17 - Prob. 17.81PCh. 17 - Prob. 17.82PCh. 17 - Prob. 17.83PCh. 17 - Prob. 17.84PCh. 17 - Prob. 17.85PCh. 17 - Prob. 17.86PCh. 17 - Prob. 17.87PCh. 17 - Prob. 17.88PCh. 17 - When 0.100 mol of CaCO3(s) and 0.100 mol of CaO(s)...Ch. 17 - Prob. 17.90PCh. 17 - Prob. 17.91PCh. 17 - Prob. 17.92PCh. 17 - Highly toxic disulfur decafluoride decomposes by a...Ch. 17 - A study of the water-gas shift reaction (see...Ch. 17 - Prob. 17.95PCh. 17 - Prob. 17.96PCh. 17 - Prob. 17.97PCh. 17 - Prob. 17.98PCh. 17 - Prob. 17.99PCh. 17 - Prob. 17.100PCh. 17 - The molecular scenes below depict the reaction Y ⇌...Ch. 17 - For the equilibrium H2S(g) ⇌ 2H2(g) + S2(g) Kc =...Ch. 17 - Prob. 17.103PCh. 17 - Prob. 17.104PCh. 17 - The kinetics and equilibrium of the decomposition...Ch. 17 - Isopentyl alcohol reacts with pure acetic acid to...Ch. 17 - Isomers Q (blue) and R (yellow) interconvert. They...Ch. 17 - Glauber’s salt, Na2SO4·10H2O, was used by J. R....Ch. 17 - Prob. 17.109PCh. 17 - Synthetic diamonds are made under conditions of...
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