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 17.100P

(a)

Interpretation Introduction

Interpretation:

Equilibrium pressure of N in N2 for the given reaction has to be calculated at 1000 K and 200 atm.

Concept Introduction:

Equilibrium constant: It is the ratio of products to reactants has a constant value when the reaction is in equilibrium at a certain temperature. And it is represented by the letter K.

For a reaction,

aA+bBcC+dD

The equilibrium constant in terms of  partial pressure is, K=(PC)c(PD)d(PA)a(PB)b

where,

a, b, c and d are the stoichiometric coefficients of reactant and product in the reactions

(a)

Expert Solution
Check Mark

Answer to Problem 17.100P

Equilibrium pressure of N is 4.0×10-21 atm.

Explanation of Solution

Given data is shown below:

  N2(g)  2N(g) logKP =43.10Initial partial pressure of N2= 200 atmTemperature = 1000 K

Equilibrium constant for the given reaction is given in terms of partial,

  Kp = (PN)2(PN2)

Calculate Kp as follows,

  log Kp = 43.10Kp = 7.943×1044

Kp for the given reaction is 7.943×1044.

Equilibrium pressure of N in N2 can be determined by constructing ICE table.

                N2(g)  2N(g)Initial2000Changex+2xEquilibrium  200x      2x     

Substitute these values for the equation of equilibrium constant as given,

  Kp = (PN)2(PN2)7.943×1044 = (2x)2(200x)

Here, assume that (200x)200. Hence,

  7.943×1044 = (2x)2(200)x = 1.99×1021

Therefore,

Equilibrium pressure of N is calculated as follows,

  PN(Equlibrium) = 2(1.99×1021)= 4.0×1021 atm N

Equilibrium pressure of N is 4.0×10-21 atm.

(b)

Interpretation Introduction

Interpretation:

Equilibrium pressure of H in H2 for the given reaction has to be calculated at 1000 K and 600 atm.

Concept Introduction:

Equilibrium constant: It is the ratio of products to reactants has a constant value when the reaction is in equilibrium at a certain temperature. And it is represented by the letter K.

For a reaction,

aA+bBcC+dD

The equilibrium constant in terms of partial pressure is, K=(PC)c(PD)d(PA)a(PB)b

where,

a, b, c and d are the stoichiometric coefficients of reactant and product in the reactions

(b)

Expert Solution
Check Mark

Answer to Problem 17.100P

Equilibrium pressure of H is 5.5×10-8 atm.

Explanation of Solution

Given data is shown below:

  H2(g)  2H(g) logKP =17.30Initial partial pressure of H2= 600 atmTemperature = 1000 K

Equilibrium constant for the given reaction is given in terms of partial,

  Kp = (PH)2(PH2)

Calculate Kp as follows,

  log Kp = 17.30Kp = 5.0119×1018

Kp for the given reaction is 5.0119×1018.

Equilibrium pressure of H in H2 can be determined by constructing ICE table.

                H2(g)  2H(g)Initial6000Changex+2xEquilibrium  600x      2x     

Substitute these values for the equation of equilibrium constant as given,

  Kp = (PH)2(PH2)5.0119×1018 = (2x)2(600x)

Here, assume that (600x)600. Hence,

  5.0119×1018 = (2x)2(600)x = 2.74×108

Therefore,

Equilibrium pressure of H is calculated as follows,

  PH(Equlibrium) = 2(2.74×108)= 5.5×108 atm H

Equilibrium pressure of H is 5.5×10-8 atm.

(c)

Interpretation Introduction

Interpretation:

Number of N atoms and H atoms for the given reaction has to be calculated.

Concept Introduction:

According to Ideal gas equation,

  Numberofmoles=Pressure×VolumeR×Temperature

Number of atoms can be determined using the given formula,

  No. of atoms = No. of moles × Avogadro's number

(c)

Expert Solution
Check Mark

Answer to Problem 17.100P

  • Number of N atoms is 29 N atoms/L.
  • Number of H atoms is 4.0×1014 H atoms/L.

Explanation of Solution

Given data is shown below:

  N2(g)  2N(g) logKP =43.10H2(g)  2H(g) logKP =17.30Equilibrium pressure of N = 4.0×10-21 atmEquilibrium pressure of H = 5.5×108 atmVolume of both reactions = 1 LTemperature of both reactions = 1000 K

  • Calculate number of moles of N:

Number of moles of N can be calculated using Ideal gas equation as follows,

PV = nRT(4.0×1021 atm)(1 L)= n(0.0821 L.atm/mol.K)(1000 K)n = (4.0×1021 atm)(1 L)(0.0821 L.atm/mol.K)(1000 K)= 4.87×1023 mol N

Number of moles of N is 4.87×1023 mol.

  • Calculate number of N atoms:

Number of N atoms can be obtained by multiplying number of moles of N with Avogadro’s number.  Number of N atoms is determined as shown below,

  No. of N atoms = (4.87×1023 N mol)(6.022×1023 N atoms1 mol N)= 29 N atoms/L 

Number of N atoms is 29 N atoms/L.

  • Calculate number of moles of H:

Number of moles of H can be calculated using Ideal gas equation as follows,

PV = nRT(5.5×108 atm)(1 L)= n(0.0821 L.atm/mol.K)(1000 K)n = (5.5×108 atm)(1 L)(0.0821 L.atm/mol.K)(1000 K)= 6.7×1010 mol H

Number of moles of H is 6.7×1010 mol H.

  • Calculate number of H atoms:

Number of H atoms can be obtained by multiplying number of moles of H with Avogadro’s number.  Number of H atoms is determined as shown below,

  No. of N atoms = (6.7×1010 mol H)(6.022×1023 N atoms1 mol N)= 4.0×1014 H atoms/L 

Number of H atoms is 4.0×1014 H atoms/L.

(c)

Interpretation Introduction

Interpretation:

Number of N atoms and H atoms for the given reaction has to be calculated.

Concept Introduction:

According to Ideal gas equation,

  Numberofmoles=Pressure×VolumeR×Temperature

Number of atoms can be determined using the given formula,

  No. of atoms = No. of moles × Avogadro's number

(c)

Expert Solution
Check Mark

Answer to Problem 17.100P

  • Number of N atoms is 29 N atoms/L.
  • Number of H atoms is 4.0×1014 H atoms/L.

Explanation of Solution

Given data is shown below:

  N2(g)  2N(g) logKP =43.10H2(g)  2H(g) logKP =17.30Equilibrium pressure of N = 4.0×10-21 atmEquilibrium pressure of H = 5.5×108 atmVolume of both reactions = 1 LTemperature of both reactions = 1000 K

  • Calculate number of moles of N:

Number of moles of N can be calculated using Ideal gas equation as follows,

PV = nRT(4.0×1021 atm)(1 L)= n(0.0821 L.atm/mol.K)(1000 K)n = (4.0×1021 atm)(1 L)(0.0821 L.atm/mol.K)(1000 K)= 4.87×1023 mol N

Number of moles of N is 4.87×1023 mol.

  • Calculate number of N atoms:

Number of N atoms can be obtained by multiplying number of moles of N with Avogadro’s number.  Number of N atoms is determined as shown below,

  No. of N atoms = (4.87×1023 N mol)(6.022×1023 N atoms1 mol N)= 29 N atoms/L 

Number of N atoms is 29 N atoms/L.

  • Calculate number of moles of H:

Number of moles of H can be calculated using Ideal gas equation as follows,

PV = nRT(5.5×108 atm)(1 L)= n(0.0821 L.atm/mol.K)(1000 K)n = (5.5×108 atm)(1 L)(0.0821 L.atm/mol.K)(1000 K)= 6.7×1010 mol H

Number of moles of H is 6.7×1010 mol H.

  • Calculate number of H atoms:

Number of H atoms can be obtained by multiplying number of moles of H with Avogadro’s number.  Number of H atoms is determined as shown below,

  No. of N atoms = (6.7×1010 mol H)(6.022×1023 N atoms1 mol N)= 4.0×1014 H atoms/L 

Number of H atoms is 4.0×1014 H atoms/L.

(d)

Interpretation Introduction

Interpretation:

More reasonable step to continue the mechanism after the catalytic dissociation has to be explained.

(d)

Expert Solution
Check Mark

Explanation of Solution

Number of N atoms present in 1.0 L is only 29 and it indicates that basically no NH molecules are produced during the first reaction.

There are more N2 molecules than N atoms in order of magnitude and hence the more reasonable step is the second reaction.

  N2(g) + H(g)  NH(g) + N(g)

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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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