
In the dissociation of $2HI(g) \rightleftharpoons {H_2}(g) + {I_2}(g)$ the degree of dissociation will be affected by:
A. Increase of temperature
B. Addition of inert gas
C. Addition of ${H_2}$ and ${I_2}$
D. increase of pressure
Answer
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Hint: We know that a dissociation constant is a specific type of equilibrium constant that measures the propensity of a larger object to separate reversibly into smaller components, as when a complex falls apart into its component molecules, or when a salt splits up into its component ions.
Complete answer:
We have to know that the degree of dissociation is the phenomenon of generating current carrying free ions, which are dissociated from the fraction of solute at a given concentration. For calculating the degree of dissociation, we know that the formula used is:
${K_a} = \dfrac{{{\alpha ^2}C}}{{(1 - \alpha )}};p{K_a} = - \log {K_a}$
When we consider dissociation in the case of $2HI(g) \rightleftharpoons {H_2}(g) + {I_2}(g)$ the degree of dissociation and the value of the equilibrium constant will be affected by increase or decrease of temperature.
Therefore, the option A is correct.
We also remember that it will not be affected by addition of inert gas, addition of ${H_2}$ and ${I_2}$ increase or decrease of pressure.
Also, by the addition of${H_2}$ and ${I_2}$ , the reaction will move towards backward direction and it will suppress the dissociation of $HI$ . But with an increase of temperature, it moves the reactions towards forward direction. Thus, it will promote the dissociation of $HI$ .
Note:
We know that the pH value is a measure of the concentration of hydrogen ions in an aqueous solution. $p{K_a}$ is the acid dissociation constant and it is related to pH value but $p{K_a}$ is more specific in that as it helps us to predict what a molecule will do at a specific pH value. Also, the relationship between pH and $p{K_a}$ can be described by the Henderson-Hasselbalch equation.
Complete answer:
We have to know that the degree of dissociation is the phenomenon of generating current carrying free ions, which are dissociated from the fraction of solute at a given concentration. For calculating the degree of dissociation, we know that the formula used is:
${K_a} = \dfrac{{{\alpha ^2}C}}{{(1 - \alpha )}};p{K_a} = - \log {K_a}$
When we consider dissociation in the case of $2HI(g) \rightleftharpoons {H_2}(g) + {I_2}(g)$ the degree of dissociation and the value of the equilibrium constant will be affected by increase or decrease of temperature.
Therefore, the option A is correct.
We also remember that it will not be affected by addition of inert gas, addition of ${H_2}$ and ${I_2}$ increase or decrease of pressure.
Also, by the addition of${H_2}$ and ${I_2}$ , the reaction will move towards backward direction and it will suppress the dissociation of $HI$ . But with an increase of temperature, it moves the reactions towards forward direction. Thus, it will promote the dissociation of $HI$ .
Note:
We know that the pH value is a measure of the concentration of hydrogen ions in an aqueous solution. $p{K_a}$ is the acid dissociation constant and it is related to pH value but $p{K_a}$ is more specific in that as it helps us to predict what a molecule will do at a specific pH value. Also, the relationship between pH and $p{K_a}$ can be described by the Henderson-Hasselbalch equation.
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