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 17.84P

(a)

Interpretation Introduction

Interpretation:

A source of ethanol is the reaction catalysed by H3PO4 of steam with ethylene derived from oil.

  C2H4(g)+ H2O(g)C2H5OH(g)ΔHr×n°= -47.8 kJ Kc= 9×103 at 600 K

At equilibrium, PC2H5OH= 200.atm and PH2O= 400.atmPC2H4 has to be calculated.

Concept Introduction:

Equilibrium constant:

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

From the given Kc value, Kp is calculated.

  Kp = Kc(RT)ΔnΔn = moles of gaseous products - moles of gaseous reactants     = 1-2 = -1Kp= (9×103)[(0.0821 L.atm/mol.K)(600.K)]-1Kp= 1.82704×102

The values are substituted in equilibrium constant expression.

  Kp = PC2H4OHPC2H4PH2O1.8270×102 = 200PC2H4(400)PC2H4= 2.7367×10-3 = 3×10-3atm

PC2H4 is 3×10-3 atm.

(b)

Interpretation Introduction

Interpretation:

The highest yield of ethanol obtained at either high or low P or at high or low T has to be given.

Concept Introduction:

Le Chatelier’s principle:

Le Chatelier’s principle states that the changes in the temperature, pressure, volume and concentration of the system results in the change in system to attain new equilibrium.  It is used to understand the conditions of a reaction which favours increased product formation.

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.

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.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is

  C2H4(g)+ H2O(g)C2H5OH(g)ΔHr×n°= -47.8 kJ Kc= 9×103 at 600 K

As the ΔHr×n° is negative, the reaction is exothermic that is heat is released from the reaction.

  C2H4(g)+ H2O(g)C2H5OH(g)+ heat

The reaction proceeds towards products at low temperature.  Heat is removed from the reaction and more yield of ethanol is obtained.

In reactants, two moles of gaseous molecules are present and in products, one mole of gaseous molecules are present.  The high pressure favours the formation of ethanol as fewer moles are present.

(c)

Interpretation Introduction

Interpretation:

A source of ethanol is the reaction catalysed by H3PO4 of steam with ethylene derived from oil.

  C2H4(g)+ H2O(g)C2H5OH(g)ΔHr×n°= -47.8 kJ Kc= 9×103 at 600 K

Kc of the reaction at 450.K has to be calculated.

(c)

Expert Solution
Check Mark

Explanation of Solution

Kc of the reaction at 450.K can be calculated as

Given,

  K1= 9×103 T1= 600.KΔHr×n° = (-47.8 kJ)(103 J1 kJ) = -4.78×104 J

Kc at 450.K is

  K2=? T2=450.K R=8.314 J/mol.K

  lnK2K1 = -ΔHr×n°R(1T2-1T1)lnK29×103 = --4.78×104 J8.314 J/mol.K(1450.K-1600.K)lnK29×103 = 3.1940769K29×103 = 24.38765K2 = (9×103)(24.38765)K2 = 2.1949×105

The value of Kc at 450.K is 2.1949×105.

(d)

Interpretation Introduction

Interpretation:

In NH3 manufacture, the yield is increased by condensing to a liquid and removing it.  The condensation of C2H5OH have the same effect in ethanol production or not has to be explained.

(d)

Expert Solution
Check Mark

Explanation of Solution

On condensation, C2H5OH yield is not increased.  It is because, the boiling point of ethanol (78.5°C) is lower than the boiling point of water (100°C).  On condensation, water molecules are removed.  So, moles of gaseous molecules will get decreased from each side of the reaction.  Hence, the equilibrium is unaffected.

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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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