Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 12, Problem 92QRT

(a)

Interpretation Introduction

Interpretation:

The mass of Ni(CO)4 that can be formed from 2.05 g CO with 0.125 g Nickel metal has to be calculated.

(a)

Expert Solution
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Explanation of Solution

Balanced reaction is,

  Ni(s)+4CO(g)Ni(CO)4(g)

Mole of CO in 2.05 g CO is,

    =2.05gCO×1molCO28.0101CO=0.0183mol

Nickel metal in 0.125 g Nickel is,

  =0.125gNi×1molNi58.6934gNi=0.000213mol

The lowest mole is limiting reagent so Nickel is the reagent.

The mass of Ni(CO)4 that can be formed from  0.000213  mol Nickel is,

  =0.000213Ni(CO)4×170.7338gNi(CO)41molNi(CO)4=0.363gNi(CO)4

Hence, the mass of Ni(CO)4 that can be formed from 2.05 g CO with 0.125 g Nickel metal is 0.363g.

(b)

Interpretation Introduction

Interpretation:

The enthalpy change of decomposition reaction of Ni(CO)4 has to be calculated.

Concept Introduction:

Hess's Law:

The enthalpy change of given reaction is calculated by subtraction of sum of enthalpy of formation reactants from sum of enthalpy of formation reactant products.

  ΔHrex=ΔHproduct-ΔHreactant

(b)

Expert Solution
Check Mark

Explanation of Solution

Balanced reaction is,

  Ni(s)+4CO(g)Ni(CO)4(g)

The standard formation enthalpy of Ni(CO)4 gas is -602.9 kJ/mol

    ΔHrex=ΔHo(Ni(s))+4ΔHo(CO(g))-ΔHo(Ni(CO)4(g)=0(kJ/mol)+4(-110.525kJ/mol)-(-602.9kJ/mol)=161.9kJ/mol

The enthalpy change of decomposition reaction of Ni(CO)4 is 161.9kJ/mol and it is an endothermic reaction.

(c)

Interpretation Introduction

Interpretation:

It has to be predicted whether there is an increase or a decrease in entropy when this reaction occurs.

  Ni(s)+4CO(g)Ni(CO)4(g)

(c)

Expert Solution
Check Mark

Explanation of Solution

Given reaction is,   

  Ni(CO)4(g)Ni(s)+4CO(g)11+4

The mole of product is increases, when entropy of the forward reaction is increased.

In the given reaction, 5 mole of products formed from 1 mole of reactant so entropy is increased in the given reaction.

(d-i)

Interpretation Introduction

Interpretation:

The equilibrium concentration of CO in the flask has to be calculated.

Concept Introduction:

Equilibrium constant(Kc):

Equilibrium constant (Kc) is the ratio of the rate constants of the forward and reverse reactions at a given temperature.  In other words it is the ratio of the concentrations of the products to concentrations of the reactants.  Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

    aAbB

    Rate of forward reaction = Rate of reverse reactionkf[A]a=kr[B]b

On rearranging,

    [B]b[A]a=kfkr=Kc

Where,

    kf is the rate constant of the forward reaction.

    kr is the rate constant of the reverse reaction.

    Kc is the equilibrium constant.

(d-i)

Expert Solution
Check Mark

Explanation of Solution

Given reaction is,

  Ni(CO)4(g)Ni(s)+4CO(g)

ICE table for at given condition with given concentrations,

    Ni(CO)4(g)Ni(s)+4CO(g)initial0.01solid0change-xsolid+4xequilibrium0.01-x4x

At equilibrium,

    [Ni(CO)4]=0.01-x=0.00001x=0.01-0.00001=0.01[CO]=4x=0.04MKc=[CO]4[Ni(CO)4]=(0.04)40.00001=0.3

Hence, the equilibrium concentration of CO in the flask is 0.04M.

(d-ii)

Interpretation Introduction

Interpretation:

The value of the equilibrium constant Kc for this reaction at 100 °C has to be calculated.

Concept Introduction:

Refer part (d-i).

(d-ii)

Expert Solution
Check Mark

Explanation of Solution

Given reaction is,

  Ni(CO)4(g)Ni(s)+4CO(g)

ICE table for at given condition with given concentrations,

    Ni(CO)4(g)Ni(s)+4CO(g)initial0.01solid0change-xsolid+4xequilibrium0.01-x4x

At equilibrium,

    [Ni(CO)4]=0.01-x=0.00001x=0.01-0.00001=0.01[CO]=4x=0.04MKc=[CO]4[Ni(CO)4]=(0.04)40.00001=0.3

Hence, the value of equilibrium constant Kc for this reaction at 100 °C is 0.3.

(d-iii)

Interpretation Introduction

Interpretation:

The value of the equilibrium constant Kp for this reaction at 100 °C has to be calculated.

(d-iii)

Expert Solution
Check Mark

Explanation of Solution

Given reaction is,

  Ni(CO)4(g)Ni(s)+4CO(g)

The relationship between Kc and Kp is,

    Kp=Kc(RT)Δn

In the given balanced reaction, mole change, temperature and Kc are,

    Δn=4molCO(g)-1molNi(CO)4(g)=3T=100°C+273=373KKc=0.3

The value of equilibrium constant Kp for this reaction at 100 °C,

    Kp=(0.3)×[(0.08206L×atmmol×K)×(373K)]3=9×103

Hence, the value of equilibrium constant Kp for this reaction at 100 °C is 9×103.

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Chapter 12 Solutions

Chemistry: The Molecular Science

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