# 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

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

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