Chemical Equilibrium | Degree of dissociation
Degree of dissociation (α)
- Fraction of reactants dissociated is known as degree of dissociation.
\[α = {{Moles\ of\ reactants\ dissociated \times 100} \over Initial\ moles\ of\ reactants}\]
Let us consider an example, in which moles of A given is 2.
\[ A ⇌ 2B + 3C\]
\[At\ t=0,\ moles\ of\ A = 2,\ moles\ of\ B = 0,\ moles\ of\ C = 0\]
At equilibrium:
\[moles\ of\ A = 2-2α,\ moles\ of\ B = 2\times 2α,\ moles\ of\ C = 3\times 3α\]
If degree of dissociation of NH3 is 75%, then calculate the partial pressure of each gas at equilibrium. The total pressure at equilibrium is 1 atm.
\(2NH_{3_{(g)}} ⇌ N_{2_{(g)}} + 3H_{2_{(g)}}\)
Let us consider that at t = 0, moles of NH3 = a, moles of N2 = 0 and moles of H2 = 0.
If α is the degree of dissociation, then at equilibrium, we can calculate:
\[moles\ of\ NH_3 = a - a\ α\]
\[moles\ of\ N_2 = {{a α} \over 2}\]
\[moles\ of\ H_2 = {{3a α} \over 2}\]
\[Total\ moles\ at\ equilibrium = a - aα + {{a α} \over 2} + {{3a α} \over 2} = a(1+α)\]
\[Partial\ pressure\ of\ NH_3 = {Χ_{NH_3} \times Total\ pressure} = {a(1-α) \over a(1+α)} \times 1 = {1-0.75 \over 1+0.75} = {1\over7} atm\]
\[Similarly,\ partial\ pressure\ of\ N_2 = {3\over14} atm\]
\[And,\ partial\ pressure\ of\ H_2 = {9\over14} atm\]
Calculations of Kc and Kp
1. Reactions with Δng = 0
Let us consider an example where Δng is zero (decomposition of HI)
\[2HI_{(g)} ⇌ H_{2_{(g)}} + I_{2_{(g)}}\]
\[At\ t=0,\ let\ moles\ of\ HI = a\]
\[At\ t=0,\ moles\ of\ H_2\ and\ I_2 = 0\]
If α is the rate of dissociation ,then at equilibrium:
\[moles\ of\ HI = a - aα\]
\[moles\ of\ H_2 = {aα\over 2}\]
\[moles\ of\ I_2 = {aα\over 2}\]
Also, let the volume of container be V. Then, the concentrations are:
\[Concentration\ of\ HI = {a(1-α)\over V}\]
\[Concentration\ of\ H_2 = {aα\over 2V}\]
\[Concentration\ of\ I_2 = {aα\over 2V}\]
Now we can calculate Kc as below:
\[K_c = {{[H_2][I_2]}\over{[HI]^2}}\]
\[K_c = {{{aα\over 2V} \times {aα\over 2V}} \over ({{a(1-α)}\over V})^2}\]
\[K_c = {α^2 \over 4(1 - α)^2}\]
Also, Kp is given by:
\[K_p = K_c\ (RT)^{Δn_g}\]
Here, Δng = 0, so we can write:
\[K_p = K_c = {α^2 \over 4(1 - α)^2}\]
Kp can also be calculated alternatively using the below method:
\[Total\ moles\ at\ equilibrium = a - aα + {aα\over 2} + {aα\over 2} = a\]
Let total pressure be P and we know that:
\[Partial\ pressure = mole\ fraction \times total\ pressure\]
Hence, we can find partial pressure of each gas as:
\[P_{H_2} = {aα\over 2a} \times P = {Pα \over 2}\]
\[P_{I_2} = {aα\over 2a} \times P = {Pα \over 2}\]
\[P_{HI} = {{a(1-α)\over a} \times P} = P(1-α)\]
\[K_p = {{P_{H_2} \times P_{I_2}} \over (P_{HI})^2} = {α^2 \over {4(1-α)^2}}\]
2. Reactions with Δng > 0
Let us consider an example where Δng > 0 (decomposition of PCl5)
\[PCl_{5_{(g)}} ⇌ PCl_{3_{(g)}} + Cl_{2_{(g)}}\]
\[At\ t=0,\ let\ moles\ of\ PCl_5 = a\]
\[At\ t=0,\ moles\ of\ PCl_3\ and\ PCl_5\ = 0\]
If α is the rate of dissociation, then at equilibrium:
\[moles\ of\ PCl_5 = a-aα\]
\[moles\ of\ PCl_3 = aα\]
\[moles\ of\ Cl_2 = aα\]
If volume of container is V, then the concentrations at equilibrium:
\[Concentration of PCl_5 = {a-aα \over V}\]
\[Concentration of PCl_3 = {aα \over V}\]
\[Concentration of Cl_2 = {aα \over V}\]
\[K_c = {[PCl_3][Cl_2]\over [PCl_5]}\]
\[K_c = {{aα\over V}\times{aα\over V}\over{a(1-α)\over V}} = {aα^2 \over (1-α)V}\]
Now, let us calculate Kp in terms of pressure. Let the total pressure be P.
\[Total\ moles\ at\ equilibrium = a-aα+aα+aα = a(1+α)\]
We know that partial pressure = mole fraction * Total Pressure. Hence:
\[P_{PCl_3} = {{aα\over a(1+α)} \times P} = {Pα\over(1+α)}\]
\[P_{Cl_2} = {{aα\over a(1+α)} \times P} = {Pα\over(1+α)}\]
\[P_{PCl_5} = {{a(1-α)\over a(1+α)}\times P} = {P(1-α)\over (1+α)}\]
We can calculate Kp as:
\[K_p = {{P_{PCl_3} \times P_{Cl_2}} \over P_{PCl_5}}\]
\[K_p = {{Pα\over (1+α)} \times {Pα\over (1+α)} \over {P(1-α)\over(1+α)}}\]
\[K_p = {Pα^2 \over (1-α^2)}\]
3. Reactions with Δng < 0
Let us consider a reaction in which Δng is less than 0.
\[N_{2_{(g)}} + 3H_{2_{(g)}} ⇌ 2NH_{3_{(g)}}\]
At time t = 0, let the moles of N2 and H2 be a and moles of NH3 be 0.
Now, if α is the degree of dissociation, then at equilibrium:
\[moles\ of\ N_2 = a - {1 \over 3}aα\]
\[moles\ of\ H_2 = a - aα\]
\[moles\ of\ NH_3 = {2\over 3} aα\]
Let the volume of container be V. Now, we can calculate Kc as shown below:
\[K_c = {[NH_3]^2 \over [N_2] [H_2]^3}\]
\[K_c = {({2aα\over 3V})^2 \over ({a-{aα\over 3} \over V}) \times ({a-aα\over V})^3 }\]
\[K_c = {4α^2V \over 3a(3-α)(1-α)^2}\]
Can you calculate Kp for this reaction in terms of total pressure P?