ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
7th Edition
ISBN: 9781319403959
Author: ATKINS
Publisher: MAC HIGHER
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Chapter 5, Problem 5.43E

(a)

Interpretation Introduction

Interpretation:

For the given reaction CO(g) + H2O(g)  CO2(g) + H2(g), when K =1, the standard Gibbs free energy of the reaction has to be calculated.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is shown below,

  CO(g) + H2O(g)  CO2(g) + H2(g)

The standard Gibbs free energy of the reaction can be expressed as follow,

  ΔGo = - RT In KK= 1ΔGo = - RT In (1)ΔGo = - RT (0)ΔGo = 0

The standard Gibbs free energy of the reaction is 0.

(b)

Interpretation Introduction

Interpretation:

For the given reaction CO(g) + H2O(g)  CO2(g) + H2(g), when K =1, the temperature of the reaction has to be calculated.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is shown below,

  CO(g) + H2O(g)  CO2(g) + H2(g)

The temperature of the reaction can be expressed as follow,

  ΔGo = - RT In KWhere,ΔGo = ΔHoTΔSoK= 1ΔGo = - RT In (1)ΔGo = - RT (0)ΔGo = 0

The standard Gibbs free energy of the reaction is 0. The enthalpy of the reaction is calculated as follows,

ΔHo=Product-ReactantΔHo=[393.51kJmol1+0kJmol1][110.53kJmol1+(241.82kJmol1)]ΔHo=41.16kJmol1

The entropy of the reaction is calculated as follows,

ΔSo=Product-ReactantΔSo=[130.68JK1mol1+213.74JK1mol1][197.67JK1mol1+188.83JK1mol1]ΔSo=42.08JK1mol1

Therefore,

  ΔGo = ΔHoTΔSo0 = ΔHoTΔSoT=ΔHoΔSoT=41.16kJmol142.08JK1mol1T=41.16kJmol1( 1 000 J . kJ-1 )42.08JK1mol1T=978K

The temperature of the reaction is 978K.

(c)

Interpretation Introduction

Interpretation:

When a cylinder is filled with CO(g) at 10.00 bar, H2O(g) at 10.00 bar, H2(g) at 5.00 bar, and CO2(g) at 5.00 bar, the partial pressure of the gases has be calculated.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is shown below,

  CO(g) + H2O(g)  CO2(g) + H2(g)

The reaction can be expressed as follow,

ICE table:

               CO(g) + H2O(g)  CO2(g) + H2(g)

Initial concentration10.00bar10.00bar5.00bar5.00bar
Changeαα+α+α
At equilibrium10.0α10.0α5.0+α5.0+α

  CO(g) + H2O(g)  CO2(g) + H2(g)α =[CO2][H2][CO][H2O]α =(5)(5)(10)(10)α =+2.50bar

The partial pressures of CO,H2O,COand H2 is given below,

  [CO] =10.0α[CO] =10.00.25bar[CO] =0.75bar

  [H2O] =10.0α[H2O] =10.00.25bar[H2O] =0.75bar

  [CO2] =5.00+α[CO2] =5.00+0.25bar[CO2] =0.75bar

  [H2] =5.00+α[H2] =5.00+0.25bar[H2] =0.75bar

Therefore, all the pressure is equal to +7.50bar

(d)

Interpretation Introduction

Interpretation:

When a cylinder is filled with CO(g) at 6.00 bar, H2O(g) at 4.00 bar, H2(g) at 5.00 bar, and CO2(g) at 10.00 bar,, the partial pressure of the gases has be calculated.

(d)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is shown below,

  CO(g) + H2O(g)  CO2(g) + H2(g)

Q value is calculated as follows,

  Q =[CO2][H2][CO][H2O]Q  =(10.0)(5.0)(6.0)(4.0)Q  =2.08

The Q value is greater than one, hence the reaction shift produce reactants.

ICE table:

               CO(g) + H2O(g)  CO2(g) + H2(g)

Initial concentration6.00bar4.00bar10.00bar5.00bar
Change+α+ααα
At equilibrium6.00+α4.0+α10.0α5.0α

Pressure,

  1 =(10.0-α)(5.0-α)(6.0+α)(4.0+α)(6.0+α)(4.0+α)=(10.0-α)(5.0-α)α2-15.0α+50.02+10.0α+24.025.0α=26.0α=1.04bar

The partial pressures of CO,H2O,COand H2 is given below,

  [CO] =6.0+α[CO] =6.0+1.04bar[CO] =7.04bar

  [H2O] =4.0+α[H2O] =4.0+1.04bar[H2O] =5.04bar

  [CO2] =10.00α[CO2] =10.001.04bar[CO2] =8.96bar

  [H2] =5.00α[H2] =5.001.04bar[H2] =3.96bar

Therefore, all the pressure is calculated.

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Chapter 5 Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

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