Chapter U6.4, Problem 5E

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation for the reaction of nitrogen with oxygen to form nitrous oxide is to be written.

Concept Introduction : Balanced chemical equation is the chemical equation in which number of atoms of every element is equal in both sides of the chemical reaction that is, in the product side and reactant side.

Answer to Problem 5E

Balanced chemical equation is,

  ${\text{2N}}_{\text{2}}(g)+{\text{O}}_{\text{2}}(g)\to 2{\text{N}}_2\text{O}\text{(}g\text{)}$

Explanation of Solution

Nitrogen gas reacts with oxygen gas to form nitrous oxide. So, the chemical equation for the reaction is as follows:

  ${\text{N}}_{\text{2}}(g)+{\text{O}}_{\text{2}}(g)\to {\text{N}}_2\text{O}\text{(}g\text{)}$

The above chemical is not a balanced chemical equation. To balance the equation, add coefficient 2 to ${\text{N}}_2\text{O}$ (nitrous oxide) and 2 to ${\text{N}}_{\text{2}}$ . So, the balanced chemical equation is,

  ${\text{2N}}_{\text{2}}(g)+{\text{O}}_{\text{2}}(g)\to 2{\text{N}}_2\text{O}\text{(}g\text{)}$

(b)

Interpretation Introduction

Interpretation: The equilibrium constant equation for the reaction of nitrogen and oxygen to form nitrous oxide is to be written.

Concept Introduction : The equilibrium constant equation for a reaction is written by using the concentration of reactants and products in equilibrium. Only the concentrations of gaseous reactants are considered in the equilibrium constant equation.

Answer to Problem 5E

Equilibrium constant equation is,

  $K=\frac{{[{\text{N}}_2\text{O}]}^2}{{{\text{[N}}_2]}^2[{\text{O}}_{\text{2}}]}$

Explanation of Solution

The balanced equation for the reaction of nitrogen and oxygen to form nitrous oxide is as follows:

  ${\text{2N}}_{\text{2}}(g)+{\text{O}}_{\text{2}}(g)\to 2{\text{N}}_2\text{O}\text{(}g\text{)}$

Now write the equilibrium constant equation.

  $K=\frac{[\text{Products}]}{\text{[Reactants}]}$

Product of the reaction is ${\text{N}}_{\text{2}}\text{O}$ and reactants are ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ .

So,

  $K=\frac{{[{\text{N}}_2\text{O}]}^2}{{{\text{[N}}_2]}^2[{\text{O}}_{\text{2}}]}$

(c)

Interpretation Introduction

Interpretation: The concentration of ${\text{O}}_{\text{2}}$ is to be calculated.

Concept Introduction : The equilibrium constant equation for a reaction is written by using the concentration of reactants and products in equilibrium. Only the concentrations of gaseous reactants are considered in the equilibrium constant equation.

Answer to Problem 5E

Concentration of ${\text{O}}_{\text{2}}$ is $0.022\text{M}$ .

Explanation of Solution

The balanced equation for the reaction of nitrogen and oxygen to form nitrous oxide is as follows:

  ${\text{2N}}_{\text{2}}(g)+{\text{O}}_{\text{2}}(g)\to 2{\text{N}}_2\text{O}\text{(}g\text{)}$

Equilibrium constant equation is,

  $K=\frac{{[{\text{N}}_2\text{O}]}^2}{{{\text{[N}}_2]}^2[{\text{O}}_{\text{2}}]}$..........(1)

Given the values are as follows:

  $\begin{array}{c}K=45\\ [{\text{N}}_{\text{2}}]=1.00\\ [{\text{N}}_{\text{2}}\text{O}]=1.00\end{array}$

Put the above values in equation (1).

  $\begin{array}{c}K=\frac{{[{\text{N}}_2\text{O}]}^2}{{{\text{[N}}_2]}^2[{\text{O}}_{\text{2}}]}\\ 45=\frac{{[1.00]}^2}{{[1.00]}^2[{\text{O}}_{\text{2}}]}\\ 45=\frac{1}{[{\text{O}}_{\text{2}}]}\\ [{\text{O}}_{\text{2}}]=0.022\text{M}\end{array}$

(d)

Interpretation Introduction

Interpretation: The indication of value of $K$ about the equilibrium mixture.

Concept Introduction: The equilibrium constant equation for a reaction is written by using the concentration of reactants and products in equilibrium. Only the concentrations of gaseous reactants are considered in the equilibrium constant equation.

Answer to Problem 5E

Value of K tells whether product or reactant is favored is in the reaction.

Explanation of Solution

The equilibrium constant K is a measure of whether starting substances or products are favored in an equilibrium mixture. A large value of $K$ indicates a large amount of product in the equilibrium mixture. A small value of K indicates very little product in the equilibrium mixture.