Question

For the equilibrium:
$LiCl.3N{H_3}(s) \to LiCl.3N{H_3}(s) + 2N{H_3}$. ${K_P} = 9at{m^2}$ at ${40^ \circ }C$. A $5 - L$ vessel contains $0.1mole$ of $LiCl.3N{H_3}$. How many moles of $N{H_3}$ should be added to the flask at this temperature to drive the backward reaction for completion?
A.Initial moles of $N{H_3} = 1.467moles$
B.Initial moles of $N{H_3} = 0.291moles$
C.Initial moles of $N{H_3} = 0.05moles$
D.Initial moles of $N{H_3} = 0.7837moles$

Answer 1

Hint: ${K_P}$ is defined as the equilibrium constant (which is defined as the ratio of the product of the pressure of all the products raised to the power of their coefficient to the product of the pressure of all the reactants raised to the power of their coefficient).

Complete step-by-step answer:Let us first read about pressure equilibrium constant ${K_P}$.
${K_P}$ is defined as the equilibrium constant (which is defined as the ratio of the product of the pressure of all the products raised to the power of their coefficient to the product of the pressure of all the reactants raised to the power of their coefficient). And while writing the expression for ${K_P}$ keep the pressure of solid reactants and products as $1$.
For example: Here in the question we are given with an equation as: $LiCl.3N{H_3}(s) \to LiCl.3N{H_3}(s) + 2N{H_3}$ so if we want to write the value of ${K_P}$ then it will be as:
${K_p} = {p_{N{H_3}}}^2$ because we have to treat the pressure of solid as one. Here if we take the moles of $N{H_3}$ as $a$ then the moles of $N{H_3}$ after the reaction will be $a + 0.2$ (because initially we take $0.1mole$ of $LiCl.3N{H_3}$).
And if we write the backward reaction of the same reaction then the relation between this ${K_P}$ and new ${K_P}$ i.e. ${K’_P}$ is as ${K’_P} = \dfrac{1}{{{K_P}}}$. So the equilibrium constant for backward reaction will be $\dfrac{1}{9}$. And if write the expression of ${K’_P}$ it will be as ${K’_P}= \dfrac{1}{{{p_{N{H_3}}}^2}}$. From here the value of ${p_{N{H_3}}}$ i.e. the pressure of $N{H_3}$ will be $3$.
Now apply the ideal gas equation which is $PV = nRT$, where $P$ is the pressure, $V$ is the volume, $n$ is number of moles, $R$ is gas constant and $T$ is the temperature.
So the value of $n = \dfrac{{PV}}{{RT}}$. Now the value of $P = 3,V = 5,R = 0.0821,T = 313$. After putting these values we get $n = 0.5837$ which is equal to $a$. As, the initial moles of $N{H_3} = a + 0.2$ so $N{H_3} = 0.7837moles$.

Hence option D i.e. initial moles of $N{H_3} = 0.7837moles$ is correct.

Note: Equilibrium constant can also be defined in terms of molar concentration (the ratio of moles to the volume) as the ratio of product of the concentration of all the products raised to the power of their coefficients to the concentration of the reactants raised to the power of their coefficient.