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 17.80P

(a)

Interpretation Introduction

Interpretation:

The given reaction is

  X2(g)+ Y2(g)2XY(g)

The reaction quotient, Q, for the given reaction has to be written.

Concept Introduction:

Reaction quotient:

The ratio of concentration terms for a reaction is known as reaction quotient.  It is denoted by Q.  For a reversible reaction of N2O4 to NO2, the reaction quotient is represented as

  N2O4(g)2NO2(g)Q = (NO2)2(N2O4)

(a)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is

  X2(g)+ Y2(g)2XY(g)

The reaction quotient for the reaction can be represented as

  Q = (XY)2(X2)(Y2)

(b)

Interpretation Introduction

Interpretation:

The given filmstrip contains five molecular scenes of a gaseous mixture as it reaches equilibrium over time.  In the filmstrip, X is purple and Y is orange.

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 17, Problem 17.80P , additional homework tip  1

Figure 1

If each particle represents 0.1 mol, Q has to be found for each scene.

Concept Introduction:

Reaction quotient:

The ratio of concentration terms for a reaction is known as reaction quotient.  It is denoted by Q.  For a reversible reaction of N2O4 to NO2, the reaction quotient is represented as

  N2O4(g)2NO2(g)Q = (NO2)2(N2O4)

(b)

Expert Solution
Check Mark

Explanation of Solution

Scene A:

X Molecules are 4, Y molecules are 4 and XY molecules are absent.  As each particle represents 0.1 mol,

  X = 4×0.1 = 0.4 molY = 4×0.1 = 0.4 mol

The Q can be represented as

  Q = (0)2(0.4)(0.4) = 0

Scene B:

X Molecules are 2, Y molecules are 2 and XY molecules are 4.  As each particle represents 0.1 mol,

   = 2×0.1 = 0.2 mol = 2×0.1 = 0.2 molXY =4×0.1  = 0.4mol

The Q can be represented as

  Q = (0.4)2(0.2)(0.2) = 4

Scene C, D, E:

X Molecules are 1, Y molecules are 1 and XY molecules are 6.  As each particle represents 0.1 mol,

  X = 1×0.1 = 0.1 molY = 1×0.1 = 0.1 molXY =6×0.1  =0.6mol

The Q can be represented as

  Q = (0.6)2(0.1)(0.1) = 36 = 4×101 mol

(c)

Interpretation Introduction

Interpretation:

The given reaction is

  X2(g)+ Y2(g)2XY(g)

If K > 1, the time progressing to the right or to the left has to be explained.

Concept Introduction:

At equilibrium, Q = K.

If Q = K, the mixture is at equilibrium.

If Q > K, the reaction proceeds towards reactants.

If Q < K, the reaction proceeds towards products.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given condition is K > 1.

Scene-A:

The value of Q is 0.  It is less than 1.  Hence, Q < K.

Scene-B:

The value of Q is 4.  It is greater than 1.  Hence, Q = K.

Scene-C, D, E:

The value of Q is 4×101.  It is greater than 1.  Hence, Q = K.

As the value of K is increasing from left to right, the reaction progresses towards right.

(d)

Interpretation Introduction

Interpretation:

The given reaction is

  X2(g)+ Y2(g)2XY(g)

The value of K has to be calculated at this temperature.

Concept Introduction:

Reaction quotient:

The ratio of concentration terms for a reaction is known as reaction quotient.  It is denoted by Q.  For a reversible reaction of N2O4 to NO2, the reaction quotient is represented as

  N2O4(g)2NO2(g)Q = (NO2)2(N2O4)

At equilibrium, Q = K.

(d)

Expert Solution
Check Mark

Explanation of Solution

As Q = K,

Q = K = 4×101.

(e)

Interpretation Introduction

Interpretation:

The given filmstrip contains five molecular scenes of a gaseous mixture as it reaches equilibrium over time.  In the filmstrip, X is purple and Y is orange.

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 17, Problem 17.80P , additional homework tip  2

Figure 1

If ΔHr×n°< 0, the scene that represents the mixture at a higher temperature has to be explained.

Concept Introduction:

Change in equilibrium due to temperature changes:

If the temperature is increased for the system, the equilibrium shifts away from the heat because of the reaction needs extra heat to use.

If the temperature is decreased for the system, the equilibrium shifts towards the heat because the heat needs to be produced to make up for the loss.

(e)

Expert Solution
Check Mark

Explanation of Solution

Among the given five scenes, the scene that represents the mixture at a higher temperature is scene B.  At higher temperatures, reaction shifts towards left forming more X2 and Y2.

(f)

Interpretation Introduction

Interpretation:

The given filmstrip contains five molecular scenes of a gaseous mixture as it reaches equilibrium over time.  In the filmstrip, X is purple and Y is orange.

Chemistry: The Molecular Nature of Matter and Change - Standalone book, Chapter 17, Problem 17.80P , additional homework tip  3

Figure 1

The scene that represents the mixture at a higher pressure (lower volume) has to be explained.

Concept Introduction:

Change in equilibrium due to pressure changes:

On increase in the system pressure, the equilibrium shifts towards fewer moles of gas, because, for gases,

  increase in pressure = decrease in volume.

On decrease in the system pressure, the equilibrium shifts towards more moles of gas, because, for gases,

  decrease in pressure = increase in volume.

(f)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is

  X2(g)+ Y2(g)2XY(g)

The gaseous molecules in reactants are 2 mols and in products are 2 mols.  As the moles is equal, the increase in pressure or decrease in volume does not affect the equilibrium.

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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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