Law of Mass Action and Homogeneous System





Law of Mass Action and Homogeneous System:

(i) Dissociation of HI:

Let ‘a’ mole of gaseous ‘HI’ be heated and let ‘x’ mole each of H_2 \text{and} I_2 be formed at the equilibrium. For the reaction,

\underset{\underset{a- 2x} a}{2 HI} \leftrightharpoons \underset{\underset{x} 0}{H_2} + \underset{\underset{x} 0}{I_2 }\underset{\underset{\hspace{10mm} \cdots (\text{at equilibrium})} \cdots (\text{Initial})}{}

 

If ‘V’ be the volume of the system, then the molar concentrations at equilibrium are:

[HI] = \dfrac{a- 2x}{V}; [H_2] = \dfrac{x}{V}; [I_2] = \dfrac{x}{V}

 

Applying the law of mass action to the system, we have

K = \dfrac{[H_2][I_2]}{[HI]^2} = \dfrac{\frac{x}{V} \times \frac{x}{V}}{\left(\frac{a- 2x}{V} \right)^2}

 

Or, K = \dfrac{x^2}{(a- 2x)^2}

This equilibrium constant, ‘K’ can also be written as k_c and in this type of reaction K_c = K_p because the total number of moles remain unchanged.

I. e., \Delta n = 0 and hence K_P = K_c (RT)^0 K_c.

 

 

(ii) Dissociation of PCI_5:

Let ‘a’ mol of PCl_5 be taken and x mol of it be dissociated at the equilibrium.

For the reaction,
\underset{\underset{a- x} a}{2 PCl_5} \leftrightharpoons \underset{\underset{x} 0}{PCl_3} + \underset{\underset{x} 0}{Cl_2 }\underset{\underset{\cdots (\text{at equilibrium})} \cdots (\text{Initial})}{}

If ‘V’ be the total volume of the system, then the molar concentrations at equilibrium are:

[PCl_5] = \dfrac{a- x}{V}; [PCl_3] = \dfrac{x}{V}; [Cl_2] = \dfrac{x}{V}

 

Applying the law of mass action, we have

 

K = \dfrac{[PCl_3][Cl_2]}{[PCl_5]^2} = \dfrac{\dfrac{x}{V} \times \dfrac{x}{V}}{\left(\dfrac{a- x}{V} \right)^2} = \dfrac{x^2}{(a- x)V}

 

If PCl_5, PCl_3 \text{and} Cl_2 are in gaseous state then,

\underset{\underset{a- x} a}{PCl_5 (g)} \hspace{10mm} \underset{\underset{x} 0}{PCl_3 (g)} + \underset{\underset{x} 0}{Cl_2 (g) }\underset{\underset{ \cdots (\text{at equilibrium})} \cdots (\text{Initial})}{}

 

Total no. of moles at equilibrium a – x + x + x = a + x Let P be the total pressure of the system then the partial pressure of the constituents is;

P_{PCl_5} = \dfrac{a- x}{a + x}.P; P_{PCl_3} = \dfrac{x}{a + x}.P; P_{Cl_2} = \dfrac{x}{a + x}.P

 

Applying the law of mass action,

K_P = \dfrac{P_{PCl_3} \times p_{Cl_2}}{P_{PCl_5}} \\[3mm] \text{or} \hspace{3mm} K_P = \dfrac{\frac{x}{a + x}.P \times \frac{x}{a + x}.P}{\frac{a- x}{a + x}.P} \\[3mm] \hspace{3mm} \text{or} K_p = \dfrac{x^2P}{(a- x)(a + x)}

 

 

(iii) Dissociation of N_2O_4:

Let us start with ‘a’ moles of N_2O_4 confined in a vessel of a volume V. Let x moles of it be dissociated at equilibrium. Then for the equation,

\underset{\underset{a- x} a}{N_2O_4} \leftrightharpoons \underset{\underset{2x} 0}{2NO_2} {\underset{\hspace{10mm} \cdots (\text{at equilibrium}} \cdots (\text{Initial)}}

molar concentrations are;

[N_2O_4] = \dfrac{a- x}{V}; [NO_2] = \left(\dfrac{2x}{V} \right)

Applying the law of mass action, we have

K = \dfrac{[NO_2]^2}{[N_2O_4]} = \dfrac{\left(\frac{2x}{V} \right)^2}{\left(\frac{a- x}{V} \right)} \\ \text{or} \hspace{3mm} K = \dfrac{4x^2}{(a- x)V}

 

 

(iv) Dissociation of NH_3:

Let us start with ‘a’ moles of NH_3 and let 2x moles of it be dissociated at the equilibrium. Let ‘V’ be the total volume of the system, then for the reaction,

\underset{\underset{a- 2x} a}{2NH_3} \leftrightharpoons \underset{\underset{x} 0}{N_2} + \underset{\underset{3x} 0}{3H_2}\underset{\underset{ \cdots (\text{at equilibrium})} \cdots (\text{Initial})}{}

 

molar concentrations are,

[NH_3] = \dfrac{a- 2x}{V}; [N_2] = \dfrac{x}{V}; [H_2] = \dfrac{3x}{V}

 

Applying the law of mass action, we have

K = \dfrac{[N_2][H_2]^3}{[NH_3]^2} = \dfrac{\frac{x}{V} \times \left(\frac{3x}{V} \right)^3}{\left(\frac{a- 2x}{V} \right)^2}

Or K = \dfrac{27x^4}{(a- 2x)^2} V^2

 

 

(v) Synthesis of NH_3:

Let us start with ‘a’ moles of N_2 and moles of H_2 in a closed vessel of volume, V. Let x moles of N_2 be used up at the equilibrium to combine with 3x moles of H_2 and let 2x moles of NH_3 be formed at the equilibrium.

Then for the reaction,

\underset{\underset{a- x} a}{N_2} \underset{\underset{b- 3X} 0}{3H_2} \leftrightharpoons \underset{\underset{2x} 0}{NH_3}\underset{\underset{\hspace{10mm} \cdots (\text{at equilibrium})}\cdots (\text{Initial})}{}

 

Molar concentrations are:

[N_2] = \dfrac{a- x}{V}; [H_2] = \dfrac{b- 3x}{V}; [NH_3] = \dfrac{2x}{V} applying the law of mass action, we have

K = \dfrac{[NH_3]^2}{[N_2][H_2]^3} = \dfrac{\left(\frac{2x}{V} \right)^2}{\left(\frac{a- x}{V} \right) \left(\frac{b- 3x}{V} \right)^3} \\[3mm] \text{or} \hspace{3mm} K = \dfrac{4x^2V^2}{(a- x)(b- 3x)^3}

Related posts:

  1. Law of Mass Action Guldberg and Waage (1884) put forward the law of mass...
  2. Law of Mass Action worksheet Law of Mass Action states that, ” The rate of...
  3. Types of Equilibrium There are two kinds of chemical equilibrium such as homogeneous...
  4. Gay-Laussac’s law Gay-Laussac’s law (1809): according to this law, “when gases react,...
  5. Dalton’s Law of Partial Pressure Dalton’s Law of Partial Pressure (1807): This law states that,...