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 17.103P

(a)

Interpretation Introduction

Interpretation:

Equilibrium pressure of N2, O2 and NO for the given reaction has to be calculated.

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 17.103P

  • Equilibrium pressure of N2 is 0.780 atm.
  • Equilibrium pressure of O2 is 0.210 atm.
  • Equilibrium pressure of NO is 2.669×10-16atm.

Explanation of Solution

Given data is shown below:

  N2(g) + O2(g)  2NO(g)Atmospheric partial pressure of N2= 0.780 atmAtmospheric partial pressure of O2= 0.210 atmKP =  4.35×1031

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

  Kp = (PNO)2(PN2)(PO2)

Equilibrium pressure of N2, O2 and NO can be determined by constructing ICE table.

                      N2(g)    +         O2(g)  2NO(g)Initial0.7800.210      0     Changexx+2xEquilibrium  (0.780x)     (0.210x)  2x

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

  Kp = (PNO)2(PN2)(PO2) 4.35×1031 = (2x)2(0.780x)(0.210x)

Here, assume that x<<K. Hence,

  4.35×1031 = (2x)2(0.780)(0.210)x = 1.3345×1016

Therefore,

Equilibrium pressure of N2, O2 and NO is calculated as follows,

  PN2(Equlibrium) = (0.7801.3345×1016)= 0.780 atm N2PO2(Equlibrium) = (0.2101.3345×1016)= 0.210 atm O2PNO(Equlibrium) = 2(1.3345×1016)= 2.669×1016atm NO

Equilibrium pressure of N2 is 0.780 atm.

Equilibrium pressure of O2 is 0.210 atm.

Equilibrium pressure of NO is 2.669×10-16atm.

(b)

Interpretation Introduction

Interpretation:

Pcontainer in the container N2(g) + O2(g)  2NO(g) has to be calculated.

Concept Introduction:

According to Dalton’s law of partial pressures, the total pressure is the sum partial pressure of all the gaseous component present.

(b)

Expert Solution
Check Mark

Answer to Problem 17.103P

Kp for the given reaction is 4.9×105.

Explanation of Solution

Given data is shown below:

Total pressure is the sum of partial pressure of N2, O2 and NO

Therefore,

Pcontainer is determined as follows,

  Pcontainer = PN2(Equlibrium)+PO2(Equlibrium)+PNO(Equlibrium)= 0.780 atm N2+ 0.210 atm O2 + 2.669×1016atm NO= 0.990 atm

Pcontainer in the container is 0.990 atm.

(c)

Interpretation Introduction

Interpretation:

KC for the given reaction N2(g) + O2(g)  2NO(g) has to be calculated.

Concept Introduction:

The relation between Kp and Kc is given by the following equation.

Kp = Kc(RT)Δngas Kp = Equilibrium constant in terms of partial pressureKc = Equilibrium constant in terms of concentrationΔngas = moles of gaseous product  moles of gaseous reactant

Only moles of gaseous products and reactants are used for calculating Δngas.

(c)

Expert Solution
Check Mark

Answer to Problem 17.103P

KC for the given reaction is 4.9×105.

Explanation of Solution

Given data is shown below:

  2NO(g) + O2(g)  2NO2(g)KP =  4.35×1031T = 298 K

  • Determine Δngas:

The total number of moles of gaseous reactants is 3 and the moles of gaseous product is 2.  Δngas is determined as follows,

  Δngas  = moles of gaseous product - moles of gaseous reactant =22 =0

  • Determine KC:

KC of the reaction can be determined from given Kp as below,

  Kp = Kc(RT)Δngas4.35×1031= Kc[(0.0821 atm.L/mol.K)(298 K)]0 Kc= 4.35×1031

Therefore,

KC for the given reaction is 4.35×10-31.

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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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