Question

The decomposition of ${F_2}$ (Molar Mass 38) shows equilibrium at 1073K as ${F_2}(g) \rightleftharpoons 2{F_{(g)}}$
At 1073K and 1 atm, the equilibrium mixture effuses 1.6 times as fast as $S{O_2}$ under similar conditions. Calculate ${K_p}$ in atm.
A) 1.20
B) 1.49
C) 2.42
D) 3.56

Answer 1

${K_p}$ is the equilibrium constant given as the ratio of the partial pressures of the products to that of the reactants. The reaction equation is required to find the value of ${K_p}$ . It is a unitless number. ${K_p} = \dfrac{{{p_{products}}}}{{{p_{reac\tan t}}}}$
According to Dalton's law the partial pressure is given as the product of the total Partial Pressure and the Mole Fraction of that substance in the solution. ${p_A} = {\chi _A}.{p_T}$

Complete Step By Step Answer:
We are given the equilibrium reaction. Consider that the initial quantity of ${F_2}$ in the solution is 1mol. And $\alpha$ be the moles of the reactant used up at equilibrium. The equilibrium can now be given as:
${F_2}(g) \rightleftharpoons 2{F_{(g)}}$

T=0	1	-
T=equilibrium	$1 - \alpha$	$2\alpha$

The total no. of moles at equilibrium $= 1 - \alpha + 2\alpha = 1 + \alpha$
We are given that the equilibrium mixture effuses 1.6 times as fast as $S{O_2}$ under similar conditions. The rate of effusion and the Molar Mass can be related by Graham's Law of diffusion which says that the Rate of effusion is inversely proportional to the root of their Molar Masses. The molar mass of $S{O_2}$ will be 64g/mol. The relation between the equilibrium mixture and $S{O_2}$ can be given as:
$\dfrac{{{r_{mix}}}}{{{r_{S{O_2}}}}} = \sqrt {\dfrac{{{M_{S{O_2}}}}}{{{M_{mix}}}}}$
$1.6 = \sqrt {\dfrac{{64}}{{{M_{mix}}}}}$
Squaring on both sides to remove the root,
$2.56 = \dfrac{{64}}{{{M_{mix}}}}$ $\to {M_{mix}} = \dfrac{{64}}{{2.56}} = 25g/mol$
The total Molar Mass of the mixture is the sum of the product of Molar Mass and the mole fraction of their constituent elements.
In this mixture we have ${F_2}$ and 2F . The mole fraction of each are $\dfrac{{1 - \alpha }}{{1 + \alpha }}\& \dfrac{{2\alpha }}{{1 + \alpha }}$ respectively.
The Molar mass of the mixture can hence can be given as: ${M_{mix}} = {\chi _{{F_2}}} \times {M_{{F_2}}} + {\chi _F} \times {M_F}$
The Molar mass of F and ${F_2}$ are 19 and 38 g/mol respectively. The molar mass of the mixture hence will be: $25 = \dfrac{{1 - \alpha }}{{1 + \alpha }} \times 38 + \dfrac{{2\alpha }}{{1 + \alpha }} \times 19$
$25(1 + \alpha ) = 30\alpha + 38 - 30\alpha$
$25 + 25\alpha = 38$
$25\alpha = 13 \to \alpha = \dfrac{{13}}{{25}} = 0.52$
The value of ${K_p} = \dfrac{{{p_{products}}}}{{{p_{reac\tan t}}}}$ , i.e. ${K_p} = \dfrac{{{{({p_F})}^2}}}{{{p_{{F_2}}}}}$
Therefore, the partial pressures of each have to be found out.
Partial Pressure of product $= {\chi _F} \times {p_T} = \dfrac{{2\alpha }}{{1 + \alpha }} \times {p_T}$
Partial pressure of reactant $= {\chi _{{F_2}}} \times {p_t} = \dfrac{{1 - \alpha }}{{1 + \alpha }} \times {p_T}$
Therefore, ${K_p} = \dfrac{{{{\left( {\dfrac{{2\alpha }}{{1 + \alpha }} \times {p_T}} \right)}^2}}}{{\dfrac{{1 - \alpha }}{{1 + \alpha }} \times {p_T}}}$
${K_p} = \dfrac{{4{\alpha ^2}{p_T}}}{{1 - {\alpha ^2}}}$
Substituting the values of alpha and ${p_T} = 1atm$ we get
${K_p} = \dfrac{{4{{(0.52)}^2} \times 1}}{{1 - {{(0.52)}^2}}} = \dfrac{{1.0016}}{{0.7296}}$
${K_p} = 1.482 \approx 1.49$
The correct answer is Option B.

Note:
In this question we are using three different concepts, one is finding the value of ${K_p}$ , Graham’s Law of diffusion- which states that more the mass of the Gas, slower it will diffuse and third is the partial pressure of any gas or solvent, which is the product of total partial pressure and mole fraction. The mole fraction of any solute or solvent $= \chi = \dfrac{{{n_{solute/solvent}}}}{{{n_{total}}}}$ where ${n_{total}} = {n_{solute}} + {n_{solvent}}$