Chapter 7, Problem 7MCP

(a)

Interpretation Introduction

Interpretation:

Equilibrium constant expression has to be written for the given reaction.

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

Explanation of Solution

Given reaction is,

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

The above equation is a balanced chemical equation.  Equilibrium constant expression for the given reaction can be written by considering the coefficients.  For the given reaction, equilibrium constant expression can be given as,

    $K_{eq}=[{\text{CO}}_{\text{2}}]$

Concentration of calcium carbonate and calcium oxide is not shown in the equilibrium constant expression as the physical state of both is solid.  This is because solids will have a fixed concentration.

(b)

Interpretation Introduction

Interpretation:

Rate law has to be written for the given reaction.

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

Explanation of Solution

Given reaction is,

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

There is only one reactants present in the given equation of chemical reaction.  Rate law can be given for the reaction as,

    $Rate=k[{\text{CaCO}}_{\text{3}}]$

(c)

Interpretation Introduction

Interpretation:

Effect of increase in temperature in the given reaction has to be determined.

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

Explanation of Solution

Given reaction is,

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

As energy is absorbed by the reactant, this is an endothermic reaction.  Increase in temperature results in the formation of more amount of product by shifting the equilibrium to the right.

(d)

Interpretation Introduction

Interpretation:

Adding more calcium carbonate shifts the position of equilibrium or not has to be given.

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

Explanation of Solution

Given reaction is,

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

Calcium carbonate is present in solid state.  Therefore, they have a fixed concentration.  Addition of calcium carbonate does not have any effect on the position of equilibrium.

(e)

Interpretation Introduction

Interpretation:

Effect of increase in pressure in the given reaction has to be determined.

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

Explanation of Solution

Given reaction is,

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

In the given equilibrium reaction, there is only gas present in the product side.  Increase in pressure cause the equilibrium to shift to the side that has few numbers of moles of gas.  Therefore, the equilibrium is shifted towards left.  Therefore, the reaction will be reactant favored.

(f)

Interpretation Introduction

Interpretation:

Effect of addition of catalyst on rate of the given reaction has to be determined.

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

Explanation of Solution

Given reaction is,

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

Addition of catalyst increases the rate of any reaction.  In the given reaction, both the forward and reverse rate is increased.  There will not be any change in the concentration of reactant or product because the rate of forward reaction is equal to the rate of reverse reaction.

(g)

Interpretation Introduction

Interpretation:

Effect of addition of catalyst on position of equilibrium of the given reaction has to be determined.

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

Explanation of Solution

Given reaction is,

    $\text{Energy}+{\text{CaCO}}_{\text{3}}(s)\rightleftharpoons \text{CaO}(s)+{\text{CO}}_{\text{2}}(g)$

Addition of catalyst increases the rate of any reaction.  In the given reaction, both the forward and reverse rate is increased.  There will not be any change in the concentration of reactant or product because the rate of forward reaction is equal to the rate of reverse reaction.  The equilibrium will be attained faster as the rates are increased but there will not be any effect on the position of equilibrium.