Chapter 18, Problem 126CP

(a)

Interpretation Introduction

Interpretation: The partial pressure of ammonia in the container needs to be determined, when the reaction is complete at constant 1.00 atm and at constant volume.

  

Concept Introduction: The ideal gas equation is the equation which gives the relations between P,V, n and T of gases. It can be written as:

  $\text{P}\times \text{V = n}\times \text{R}\times \text{T}$

Usually gases do not follow ideal behavior under all conditions of pressure, temperature and volumes. The relation of P, V, n and T of real gases can be given by Vander Waals equation:

  $\left(\text{P + }\frac{{\text{n}}^{\text{2}}\text{a}}{{\text{V}}^{\text{2}}}\right)\left(\text{V- n b}\right)\text{ = nRT}$

Hence $\text{P = }\frac{\text{nRT}}{\text{ V-nb}}\text{ - }\frac{{\text{n}}^{\text{2}}\text{a}}{{\text{V}}^{\text{2}}}$

Answer to Problem 126CP

  ${\text{Partial pressure of NH}}_{\text{3}}\text{= 0}\text{.5 atm}$

Explanation of Solution

Number of ${\text{H}}_{\text{2}}$ molecule = 6 molecules

Number of ${\text{N}}_{\text{2}}$ gas = 6 molecules

The balance chemical equation for the formation of ammonia:

  ${\text{N}}_{\text{2}}+{\text{ 3 H}}_{\text{2}}\stackrel{}{\to }{\text{2 NH}}_3$

Hence 1 molecule of ${\text{N}}_{\text{2}}$ reacts with 3 ${\text{H}}_{\text{2}}$ molecule to form 2 molecule of ${\text{NH}}_3$ .

Hence number of ${\text{N}}_{\text{2}}$ reacts with 6 molecules of ${\text{H}}_{\text{2}}$ :

  ${\text{6 molecules of H}}_{\text{2}}\text{×}\frac{{\text{1 molecules of N}}_{\text{2}}}{{\text{3 molecule of H}}_{\text{2}}}{\text{=2 molecules of N}}_{\text{2}}$

Thus remaining ${\text{N}}_{\text{2}}$ gas molecule = 6 − 2 = 4 molecules

Number of ammonia molecule formed = 2 x 2 = 4 molecules.

Total molecules at completion of reaction = 4 +4 = 16 molecules

Calculate mole fraction of ammonia:

  ${\text{Mole fraction of NH}}_{\text{3}}\text{=}\frac{\text{4 molecule}}{\text{(4+4)molecules}}\text{= 0}\text{.5 }$

Calculate partial pressure of ${\text{NH}}_3$ :

  $\begin{array}{l}{\text{Partial pressure of NH}}_{\text{3}}{\text{=P}}_{\text{tota}}{}_{\text{l}}\text{× mole fraction}\\ {\text{Partial pressure of NH}}_{\text{3}}\text{= 1}\text{.0 atm × 0}\text{.5}\\ {\text{Partial pressure of NH}}_{\text{3}}\text{= 0}\text{.5 atm}\end{array}$

(b)

Interpretation Introduction

Interpretation: Interpret the mole fraction of ammonia in the container when the reaction is complete at constant 1.00 atm and at constant volume.

  

Concept Introduction: The ideal gas equation is the equation which gives the relations between P,V, n and T of gases. It can be written as:

  $\text{P}\times \text{V = n}\times \text{R}\times \text{T}$

Usually gases do not follow ideal behavior under all conditions of pressure, temperature and volumes. The relation of P, V, n and T of real gases can be given by Vander Waals equation:

  $\left(\text{P + }\frac{{\text{n}}^{\text{2}}\text{a}}{{\text{V}}^{\text{2}}}\right)\left(\text{V- n b}\right)\text{ = nRT}$

Hence $\text{P = }\frac{\text{nRT}}{\text{ V-nb}}\text{ - }\frac{{\text{n}}^{\text{2}}\text{a}}{{\text{V}}^{\text{2}}}$

Answer to Problem 126CP

  ${\text{Mole fraction of NH}}_{\text{3}}\text{= 0}\text{.5 }$

Explanation of Solution

Number of ${\text{H}}_{\text{2}}$ molecule = 6 molecules

Number of ${\text{N}}_{\text{2}}$ gas = 6 molecules

The balance chemical equation for the formation of ammonia:

  ${\text{N}}_{\text{2}}+{\text{ 3 H}}_{\text{2}}\stackrel{}{\to }{\text{2 NH}}_3$

Hence 1 molecule of ${\text{N}}_{\text{2}}$ reacts with 3 ${\text{H}}_{\text{2}}$ molecule to form 2 molecule of ${\text{NH}}_3$ .

Hence number of ${\text{N}}_{\text{2}}$ reacts with 6 molecules of ${\text{H}}_{\text{2}}$ :

  ${\text{6 molecules of H}}_{\text{2}}\text{×}\frac{{\text{1 molecules of N}}_{\text{2}}}{{\text{3 molecule of H}}_{\text{2}}}{\text{=2 molecules of N}}_{\text{2}}$

Thus remaining ${\text{N}}_{\text{2}}$ gas molecule = 6 − 2 = 4 molecules

Number of ammonia molecule formed = 2 x 2 = 4 molecules.

Total molecules at completion of reaction = 4 +4 = 16 molecules

Calculate mole fraction of ammonia:

  ${\text{Mole fraction of NH}}_{\text{3}}\text{=}\frac{\text{4 molecule}}{\text{(4+4)molecules}}\text{= 0}\text{.5 }$

(c)

Interpretation Introduction

Interpretation:Interpret the volume of the container when the reaction complete if the initial pressure is 1.0 atm and volume is 15.0 L. The partial pressure of ammonia is 0.5 atm when reaction is complete.

  

Concept Introduction: The ideal gas equation is the equation which gives the relations between P,V, n and T of gases. It can be written as:

  $\text{P}\times \text{V = n}\times \text{R}\times \text{T}$

Usually gases do not follow ideal behavior under all conditions of pressure, temperature and volumes. The relation of P, V, n and T of real gases can be given by Vander Waals equation:

  $\left(\text{P + }\frac{{\text{n}}^{\text{2}}\text{a}}{{\text{V}}^{\text{2}}}\right)\left(\text{V- n b}\right)\text{ = nRT}$

Hence $\text{P = }\frac{\text{nRT}}{\text{ V-nb}}\text{ - }\frac{{\text{n}}^{\text{2}}\text{a}}{{\text{V}}^{\text{2}}}$

Answer to Problem 126CP

Volume = 10.0 L

Explanation of Solution

Initial volume is 15 L. Now, total initial moles are 2n. The final number of moles can be calculated by taking sum of moles of ammonia and moles of nitrogen.

  $n_f=\frac{2n}{3}+\frac{2n}{3}=\frac{4n}{3}\text{ mol}$

At constant temperature, pressure can be calculated for an ideal gas as follows:

  $\frac{V_i}{n_i}=\frac{V_f}{n_f}$

Putting the values,

  $\begin{array}{l}\frac{15\text{ L}}{2n}=\frac{V_f}{\frac{4n}{3}}\\ V_f=\text{10 L}\end{array}$

Thus, the final volume is 10 L.