Chapter 8, Problem 162CP

a)

Interpretation Introduction

Interpretation: The partial pressure of ammonia after the completion of given reaction need to be determined.

Concept introduction:

• Balanced equation of a reaction is written according to law of conservation of mass.

• According toideal gas law: the pressure of gas is directly proportional to the number of moles of gas at constant volume and temperature.

$\frac{\text{P}}{\text{n}}\text{=}\frac{\text{RT}}{\text{V}}$ , P and n are proportional.

Mathematically, for two gases (1 and 2) at constant V and T,

$\frac{{\text{P}}_{\text{1}}}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}}{{\text{n}}_{\text{2}}}\Rightarrow \frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}$

• According toideal gas law, the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

$\frac{\text{V}}{\text{n}}\text{=}\frac{\text{RT}}{\text{P}}$ , V and n are proportional.

Mathematically, for two gas mixtures (1 and 2) at constant P and T,

$\frac{{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}}\Rightarrow \frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}\text{=}\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}$

• Mole ratios between the reactant and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.

• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

• Total pressure of a flask containing mixture of gases is the sum of individual partial pressures of constituted gases.

To determine: the partial pressure of ammonia after the completion of given reaction.

Answer to Problem 162CP

The partial pressure of ${\text{NH}}_3$ in the container is $\text{1}\text{.00}\text{atm}$.

Explanation of Solution

To find: the balanced chemical equation of the given reaction.

${\text{N}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\to \text{2}{\text{NH}}_{\text{3}}$

The reactants of the given reaction are ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ gases.

The product of the given reaction is ${\text{NH}}_{\text{3}}$.

Therefore,

The chemical equation of the reaction given is,

${\text{N}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\to {\text{NH}}_{\text{3}}$

This equation is not balanced, Balanced equation is the chemical equation of a reaction which is written according to law of conservation of mass the mass of reactant molecules should conserve in product molecules.

The balanced chemical equation of the given reaction is,

Balance the molecules in both sides of arrow in order of atoms other than hydrogen then hydrogen atoms.

Therefore, the balanced chemical equation of the given reaction will be,

${\text{N}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\to \text{2}{\text{NH}}_{\text{3}}$

Hence,

The balanced chemical equation for the given reaction is, ${\text{N}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\to \text{2}{\text{NH}}_{\text{3}}$.

To find: the limiting reagent in the given reaction of ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$.

${\text{H}}_{\text{2}}$ is the limiting reagent in the given reaction.

The partial pressures of each reactant gases are given as $\text{1}\text{.00}\text{atm}$.

Therefore,

The partial pressure of ${\text{N}}_{\text{2}}$ is $\text{1}\text{.00}\text{atm}$.

The partial pressure of ${\text{H}}_{\text{2}}$ is $\text{1}\text{.00}\text{atm}$.

Initially the volume and temperature is constant.

According to ideal gas law, the pressure of gas is directly proportional to the number of moles of gas at constant volume and temperature.

$\frac{\text{P}}{\text{n}}\text{=}\frac{\text{RT}}{\text{V}}$ , P and n are proportional.

Mathematically, for two gases (1 and 2) at constant V and T,

$\frac{{\text{P}}_{\text{1}}}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}}{{\text{n}}_{\text{2}}}\Rightarrow \frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}$

The partial pressures ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ are equal.

Therefore,

The number of moles ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ should be equal.

The balanced chemical equation for the given reaction is, ${\text{N}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\to \text{2}{\text{NH}}_{\text{3}}$.

Here, 1 mole of ${\text{N}}_{\text{2}}$ is reacting with 3 moles of ${\text{H}}_{\text{2}}$. But in the given reaction the number of moles ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ are equal, so the ${\text{H}}_{\text{2}}$ will finish first.

Limiting reagent: limiting reagent is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction.

Hence,

${\text{H}}_{\text{2}}$ is the limiting reagent in the given reaction.

#assumes the number of moles of ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ is as x.

To find: the number of moles produced ${\text{NH}}_3$, reacted ${\text{N}}_{\text{2}}$ andremaining ${\text{N}}_{\text{2}}$ in the given reaction.

The number of moles produced ${\text{NH}}_3$ is $\frac{\text{2x}}{\text{3}}\text{mol}$.

The number of moles reacted ${\text{N}}_{\text{2}}$ is $\frac{\text{x}}{\text{3}}\text{mol}$.

The number of moles remaining ${\text{N}}_{\text{2}}$ is $\frac{\text{2x}}{\text{3}}\text{mol}$.

The limiting reagent in the given reaction is ${\text{H}}_{\text{2}}$.

The balanced equation of given reaction is ${\text{N}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\to \text{2}{\text{NH}}_{\text{3}}$.

The mole ratio between ${\text{NH}}_3$ and ${\text{H}}_{\text{2}}$ is $\text{2}\text{:}\text{3}\text{or}\text{1:}\frac{\text{3}}{\text{2}}$.

If the number of moles of ${\text{H}}_{\text{2}}$ is x, then the number of moles produced ${\text{NH}}_3$ is,

$\text{x}\text{mol}{\text{H}}_{\text{2}}\text{×}\frac{\text{2}\text{mol}{\text{NH}}_{\text{3}}}{\text{3}\text{mol}{\text{H}}_{\text{2}}}\text{=}\frac{\text{2x}}{\text{3}}\text{mol}{\text{NH}}_{\text{3}}$

The mole ratio between reacted ${\text{N}}_{\text{2}}$ and ${\text{H}}_{\text{2}}$ is $\text{1}\text{:}\text{3}$.

If the number of moles of ${\text{H}}_{\text{2}}$ is x, then the number of moles reacted ${\text{N}}_{\text{2}}$ is,

$\text{x}\text{mol}{\text{H}}_{\text{2}}\text{×}\frac{1\text{mol}{\text{N}}_{\text{2}}}{\text{3}\text{mol}{\text{H}}_{\text{2}}}\text{=}\frac{\text{x}}{\text{3}}\text{mol}{\text{N}}_{\text{2}}$

The number of moles of taken ${\text{N}}_{\text{2}}$ is x mole and the number of moles reacted ${\text{N}}_{\text{2}}$ is $\frac{\text{x}}{\text{3}}\text{mol}$.

Therefore,

The number of moles remaining ${\text{N}}_{\text{2}}$ is $\text{x}\text{mol}-\frac{\text{x}}{\text{3}}\text{mol}\text{=}\frac{\text{2x}}{\text{3}}\text{mol}$.

Hence,

The number of moles produced ${\text{NH}}_3$ is $\frac{\text{2x}}{\text{3}}\text{mol}$.

The number of moles reacted ${\text{N}}_{\text{2}}$ is $\frac{\text{x}}{\text{3}}\text{mol}$.

The number of moles remaining ${\text{N}}_{\text{2}}$ is $\frac{\text{2x}}{\text{3}}\text{mol}$.

To determine: the partial pressure of ammonia after the completion of given reaction.

The partial pressure of ${\text{NH}}_3$ in the container is $\text{1}\text{.00}\text{atm}$.

The initial total pressure of the container is given as $\text{2}\text{atm}$. Since the pressure is constant in the entire reaction process, so after the completion of reaction also the pressure of the container will be $\text{2}\text{atm}$.

The gaseous molecules presented after the completion of reaction are remaining ${\text{N}}_{\text{2}}$ gas and produced ${\text{NH}}_3$.

The number of moles produced ${\text{NH}}_3$ is calculated as $\frac{\text{2x}}{\text{3}}\text{mol}$.

The number of moles remaining ${\text{N}}_{\text{2}}$ is calculated as $\frac{\text{2x}}{\text{3}}\text{mol}$.

So after the completion of reaction, the volume and temperature is constant.

According to ideal gas law, the pressure of gas is directly proportional to the number of moles of gas at constant volume and temperature.

$\frac{\text{P}}{\text{n}}\text{=}\frac{\text{RT}}{\text{V}}$ , P and n are proportional.

Mathematically, for two gases (1 and 2) at constant V and T,

$\frac{{\text{P}}_{\text{1}}}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}}{{\text{n}}_{\text{2}}}\Rightarrow \frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}$

Therefore,

The partial pressures of ${\text{NH}}_3$ and remaining ${\text{N}}_{\text{2}}$ should be equal.

The total pressure of the container is given as $\text{2}\text{atm}$.

That means,

${\text{P}}_{{\text{NH}}_{\text{3}}}\text{+}{\text{P}}_{{\text{N}}_{\text{2}}}\text{=}\text{2}\text{atm}\Rightarrow {\text{P}}_{{\text{NH}}_{\text{3}}}\text{=}{\text{P}}_{{\text{N}}_{\text{2}}}=1\text{atm}$

Hence,

The partial pressure of ${\text{NH}}_3$ in the container is $\text{1}\text{.00}\text{atm}$.

b)

Interpretation Introduction

Interpretation: The volume of the given container after the completion of given reaction are needed to be determined.

Concept introduction:

• Balanced equation of a reaction is written according to law of conservation of mass.

• According toideal gas law: the pressure of gas is directly proportional to the number of moles of gas at constant volume and temperature.

$\frac{\text{P}}{\text{n}}\text{=}\frac{\text{RT}}{\text{V}}$ , P and n are proportional.

Mathematically, for two gases (1 and 2) at constant V and T,

$\frac{{\text{P}}_{\text{1}}}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{P}}_{\text{2}}}{{\text{n}}_{\text{2}}}\Rightarrow \frac{{\text{P}}_{\text{1}}}{{\text{P}}_{\text{2}}}\text{=}\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}$

• According toideal gas law, the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

$\frac{\text{V}}{\text{n}}\text{=}\frac{\text{RT}}{\text{P}}$ , V and n are proportional.

Mathematically, for two gas mixtures (1 and 2) at constant P and T,

$\frac{{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}}\Rightarrow \frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}\text{=}\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}$

• Mole ratios between the reactant and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.

• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

• Total pressure of a flask containing mixture of gases is the sum of individual partial pressures of constituted gases.

To determine: the volume of the given container after the completion of given reaction.

Answer to Problem 162CP

The volume of the given container after the completion of given reaction is $\text{10}\text{L}$.

Explanation of Solution

To determine: the volume of the given container after the completion of given reaction.

The volume of the given container after the completion of given reaction is $\text{10}\text{L}$.

The number of moles produced ${\text{NH}}_3$ is calculated as $\frac{\text{2x}}{\text{3}}\text{mol}$.

The number of moles remaining ${\text{N}}_{\text{2}}$ is calculated as $\frac{\text{2x}}{\text{3}}\text{mol}$.

Therefore,

The total number of moles after the completion of given reaction is $\frac{\text{2x}}{\text{3}}\text{mol}\text{+}\frac{\text{2x}}{\text{3}}\text{mol}\text{=}\frac{\text{4x}}{\text{3}}\text{mol}$.

The initial number of moles is taken as ${\text{n}}_{{\text{N}}_{\text{2}}}\text{+}{\text{n}}_{{\text{H}}_{\text{2}}}\text{=}\text{x+x}\text{=}\text{2x}$.

The initial volume of container is given as $\text{15}\text{L}$.

The reaction is carried under constant pressure and temperature.

According to ideal gas law, the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

$\frac{\text{V}}{\text{n}}\text{=}\frac{\text{RT}}{\text{P}}$ , V and n are proportional.

Mathematically, for two gas mixtures (1 and 2) at constant P and T,

$\frac{{\text{V}}_{\text{1}}}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{V}}_{\text{2}}}{{\text{n}}_{\text{2}}}\Rightarrow \frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}\text{=}\frac{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}$

Therefore,

• The relationship between final volume of the container to initial volume of container is,

  $\frac{\text{15}\text{L}}{{\text{V}}_{\text{initial}}}\text{=}\frac{\frac{\text{4x}}{\text{3}}\text{mol}}{\text{2}\text{x}}\text{=}\frac{\text{2}}{\text{3}}$

That means,

The volume of the given container after the completion of given reaction is,

$\frac{\text{15}\text{L}\times \text{2}}{3}=10\text{L}$.

Hence, the volume of the given container after the completion of given reaction is $\text{10}\text{L}$.