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Question: For the equilibrium, , what is the temperature at which \[\dfrac{{{k_p}\left( {atm} \right)}}{{{k_c}...

For the equilibrium, , what is the temperature at which kp(atm)kc(M)=3\dfrac{{{k_p}\left( {atm} \right)}}{{{k_c}\left( M \right)}} = 3?
A.0.027$$$$K
B.0.36$$$$K
C.36.54$$$$K
D.273$$$$K

Explanation

Solution

We should know if the reaction needs to be balanced in such cases. Also, the terms kp,kc{k_p},{k_c} should be clear to us and their relation with the number of moles of the gaseous components.

Formula used:
kpkc=(RT)Δn(g)\dfrac{{{k_p}}}{{{k_c}}} = {\left( {RT} \right)^{\Delta n(g)}}

Complete step by step answer:
Before approaching the solution, we must understand the basics required to tackle the question.
Consider the following reaction:
aA + bB→cC + dD\text{aA + bB} \to \text{cC + dD}
A and B are the reactants. C and D are the products formed. a,b,c,da,b,c,d represents the number of moles of the components. is the symbol showing equilibrium.
In a reversible chemical reaction, Equilibrium constant is defined as the value that expresses the relationship between the amount (in terms of pressure, concentration or molarity) of components (products and reactants) existing in equilibrium at a given temperature.
For equation (1)
Equilibrium constant in terms of partial pressure, kp=(C)c(D)d(A)a(B)b{k_p} = \dfrac{{{{\left( C \right)}^c}{{\left( D \right)}^d}}}{{{{\left( A \right)}^a}{{\left( B \right)}^b}}}
Equilibrium constant in terms of molarity, kc=[C]c[D]d[A]a[B]b{k_c} = \dfrac{{{{\left[ C \right]}^c}{{\left[ D \right]}^d}}}{{{{\left[ A \right]}^a}{{\left[ B \right]}^b}}}
Now, relation between kp,kc{k_p},{k_c} is defined as:
kp=kc(RT)Δn(g){k_p} = {k_c}{\left( {RT} \right)^{\Delta n(g)}}where R=universal gas constant= 8.314Jmol−1K−18.314Jmo{l^{ - 1}}{K^{ - 1}} -(2)
T=Temperature
Δn(g)\Delta n(g)=number of moles of gaseous components on product side- number of moles of gaseous components on reactant side
Δn(g)=(c+d)−(b+a)\Delta n(g) = (c + d) - (b + a)
Coming back to question;

Δn(g)=2−1=1\Delta n(g) = 2 - 1 = 1
Using equation (2)
kpkc=(RT)Δn(g)\dfrac{{{k_p}}}{{{k_c}}} = {\left( {RT} \right)^{\Delta n(g)}}and substituting all the values,
⇒3=(8.314×T)1\Rightarrow 3 = {\left( {8.314 \times T} \right)^1}
Thus,T=0.36KT = 0.36K

Thus option (b) is the correct option.

Note:
We need to take gaseous components into consideration in such cases. This will help us to calculate the value of difference in moles on the reactant and product side with higher accuracy.