Question

True relationship for the following reaction is \[\log \dfrac{{{K_P}}}{{{K_C}}} + \log RT = 0\]:
A. $PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
B. $2S{O_2} + {O_2} \rightleftharpoons 2S{O_3}$
C. ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$
D. $\left( 2 \right)$ and $\left( 3 \right)$ both

Answer 1

Hint:
The process of finding out the specific equation as per the given condition is to determine the $\Delta n$ of the reaction. The change in the number of moles of product from that of the reactant is used to determine the possible value of $\Delta n$. Based on the generalised equation the process of reactant to product formation can be written as:
$aA + bB \rightleftharpoons cC + dD$
This reaction suggests that the $a$ moles of the $A$ molecules and $b$ moles of the $B$ molecule are the reactants for the process. There are $c$ moles of the molecule $C$ and $d$ moles of the molecule $D$ which are formed as the product of the reaction.

Complete step by step answer:
The equation for equilibrium constants of the reaction suggests that:
$\dfrac{{{K_p}}}{{{K_C}}} = {\left( {RT} \right)^{\Delta n}}$
From here we can get the $\dfrac{{{K_P}}}{{{{\left( {RT} \right)}^{\Delta n}}}} = {K_C}$
Here the value for $\Delta n$ needs to be found out and in the process according to generalised equation:
$\Delta n = \left( {c + d} \right) - \left( {a + b} \right)$ Since it is the difference between products and the reactants.
$R$ is the gas constant in the equation for the ideal gas law. $T$ is the temperature of the reaction which is given in Kelvin. Here the given form of the reaction is:
$\log \dfrac{{{K_p}}}{{{K_C}}} + \log RT = 0$
Putting the value of ${K_C}$ here we get:
$ \Rightarrow \log \dfrac{{{K_P}}}{{\dfrac{{{K_P}}}{{{{\left( {RT} \right)}^{\Delta n}}}}}} + \log RT = 0$
From here we get,$ \Rightarrow \log {\left( {RT} \right)^{\Delta n}} + \log RT = 0$
Getting the logarithms into multiplication form, we get,
$ \Rightarrow \log \left\{ {{{\left( {RT} \right)}^{\Delta n}} \times RT} \right\} = 0$
Therefore, based on the indices formula,
$ \Rightarrow \log {\left( {RT} \right)^{\Delta n + 1}} = 0$
Removing the log from two sides we get:
$ \Rightarrow {\left( {RT} \right)^{\Delta n + 1}} = 0$
Therefore, from here we can say,
$ \Rightarrow \Delta n + 1 = 0$ which leads to the value $\Delta n = - 1$
Therefore, for the given condition based on the reaction, the value of $\Delta n = - 1$.
In the first reaction the value of $\Delta n = 2 - 1 = 1$
In the second process of the reaction the value of $\Delta n = 2 - 3 = - 1$
The third reaction process has the value of $\Delta n = 2 - 4 = - 2$
Therefore, the second reaction fits the physical condition as given in the reaction process as the given value is $\Delta n = - 1$.
Hence the second reaction has the physical parameters as given in the equation. Therefore, the correct option here is B. $2S{O_2} + {O_2} \rightleftharpoons 2S{O_3}$.

Note: For proving the physical parameters to be true the value of $\Delta n$ needs to be determined for the equation as well as the parameter given. The number of moles of the molecules involved in the reaction determines the values of the equilibrium constant of the reaction.