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Hint: To answer this question, you should recall the concept of dissociation. Calculate the change in the number of atoms of the constituents in the reaction and use the relation between \[{{\text{K}}_{\text{p}}}\] and \[{{\text{K}}_{\text{c}}}\] to find the answer to this question.
Formula used:
\[\alpha = \sqrt {\dfrac{{{K_a}}}{C}} \] where, \[\alpha \]=Degree of dissociation, \[{K_a}\]=Dissociation constant and ${C_{}}$= concentration of electrolyte
Complete step by step answer:
For any weak electrolyte, the dissociation can be written as
$HA \to {H^ + } + {A^ - }$.
The given values in the question are \[\alpha \]=\[0.0114\], ${C_1}$=$0.1{\text{M}}$, \[{C_2}\]=$0.001{\text{M}}$. For both the cases we substitute the known values in the equation of degree of dissociation:
Case 1: \[0.0114 = \sqrt {\dfrac{{{K_a}}}{{0.1}}} \]
$ \Rightarrow {{\text{K}}_{\text{a}}} = 0.0000129$
Case 2: \[\alpha = \sqrt {\dfrac{{{K_a}}}{{0.001}}} \]
$ \Rightarrow \alpha = 0.114$
On dividing these both values and solving, we get:
\[\alpha = 0.114\]
Note:
The change in values of concentration, pressure, catalyst, inert gas addition, etc. not affect equilibrium constant. The Le Chatelier's principle states that the temperature, concentration, pressure, catalyst, inert gas addition to a chemical reaction can lead to a shift in equilibrium position only. We know that activation energy is the minimum energy required to start a chemical reaction. Collisions of particles lead to reactions. Only particles that collide sufficiently, can react. Now comes the important point. From the kinetic and collision model it can be said that the rate of a reaction increases with increase in temperature due to more energy and more collisions. However, this increase in the rate of reaction depends on the value of energy of activation of the reaction which is different for both the forward and the backward reaction. So, a given increase in temperature leads to an increase not only in the rate of forward reaction but also the backward reactions to different extent. So, this proves that the value of the equilibrium constant changes with temperature only.
Formula used:
\[\alpha = \sqrt {\dfrac{{{K_a}}}{C}} \] where, \[\alpha \]=Degree of dissociation, \[{K_a}\]=Dissociation constant and ${C_{}}$= concentration of electrolyte
Complete step by step answer:
For any weak electrolyte, the dissociation can be written as
$HA \to {H^ + } + {A^ - }$.
The given values in the question are \[\alpha \]=\[0.0114\], ${C_1}$=$0.1{\text{M}}$, \[{C_2}\]=$0.001{\text{M}}$. For both the cases we substitute the known values in the equation of degree of dissociation:
Case 1: \[0.0114 = \sqrt {\dfrac{{{K_a}}}{{0.1}}} \]
$ \Rightarrow {{\text{K}}_{\text{a}}} = 0.0000129$
Case 2: \[\alpha = \sqrt {\dfrac{{{K_a}}}{{0.001}}} \]
$ \Rightarrow \alpha = 0.114$
On dividing these both values and solving, we get:
\[\alpha = 0.114\]
Note:
The change in values of concentration, pressure, catalyst, inert gas addition, etc. not affect equilibrium constant. The Le Chatelier's principle states that the temperature, concentration, pressure, catalyst, inert gas addition to a chemical reaction can lead to a shift in equilibrium position only. We know that activation energy is the minimum energy required to start a chemical reaction. Collisions of particles lead to reactions. Only particles that collide sufficiently, can react. Now comes the important point. From the kinetic and collision model it can be said that the rate of a reaction increases with increase in temperature due to more energy and more collisions. However, this increase in the rate of reaction depends on the value of energy of activation of the reaction which is different for both the forward and the backward reaction. So, a given increase in temperature leads to an increase not only in the rate of forward reaction but also the backward reactions to different extent. So, this proves that the value of the equilibrium constant changes with temperature only.
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