
At a certain temperature, only $50%$ $HI$ is dissociated into ${{H}_{2}}$ and ${{I}_{2}}$ at equilibrium. The equilibrium constant is:
A. $1.0$
B. $3.0$
C. $0.5$
D. $0.25$
Answer
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Hint: Dissociation constant is the ratio of the concentration of product to concentration of reactant. Degree of dissociation is only used in decomposition reaction. If initial moles are not given in a chemical reaction then the value of degree of dissociation is one
Formula used: $\alpha =\dfrac{D.M}{I.M}$
Where, $\alpha $ is degree of dissociation
$D.M$ is dissociated moles
$I.M$ is initial moles
$K=\dfrac{\left[ P \right]}{\left[ R \right]}$
$K$ is the dissociation constant
$\left[ P \right]$ is the concentration of product
$\left[ R \right]$ is the concentration of reactant
Complete step by step answer:
Here, it is given that the $HI$ is dissociated $50%$
Degree of dissociation $\left( \alpha \right)=\dfrac{D.M}{I.M}$
$\alpha $ is degree of dissociation
$D.M$ is dissociated moles
$I.M$ is initial moles
Now, substituting the value we get,
$\alpha =\dfrac{50}{100}$
$\alpha =0.5$
Now, let us see the reaction
$2H{{I}_{(g)}}\rightleftharpoons {{H}_{2(g)}}+{{I}_{2(g)}}$
To calculate dissociation constant $\left( K \right)$ the formula used is:
$K=\dfrac{\left[ P \right]}{\left[ R \right]}$
Where, $K$ is the dissociation constant
$\left[ P \right]$ is the concentration of product
$\left[ R \right]$ is the concentration of reactant
now, we will substitute the values in above formula we get,
At equilibrium, $K=\dfrac{{{\alpha }^{2}}}{{{\left( 2-2\alpha \right)}^{2}}}$
As we have calculated above the degree of dissociation is $0.5$
Now if we substitute the values in above formula we get,
$K=\dfrac{{{\left( 0.5 \right)}^{2}}}{{{\left( 2-2\times 0.5 \right)}^{2}}}$
On further solving,
$K=\dfrac{0.25}{{{1}^{2}}}$
$K=0.25$
So, the correct answer is “Option D”.
Additional information
The equilibrium constant is defined as the ratio of the concentration of reactants and products.
There are several characteristics of equilibrium constant:
The catalyst changes the rate of reaction but it does not change the value of equilibrium constant.
If the equilibrium is reversed then the value of equilibrium constant is reciprocal of the original equilibrium constant.
If the reaction is multiplied by the constant value $n$ then the value of equilibrium constant becomes the ${{n}^{th}}$ power of the original value.
When two or more reactions are added then the final reaction becomes the product of individual equilibrium constant.
Note: Degree of dissociation is defined as the ratio of dissociated moles to the initial moles.
Equilibrium constant is represented in many forms like equilibrium constant at constant pressure, equilibrium constant at constant concentration, equilibrium constant at constant moles.
Formula used: $\alpha =\dfrac{D.M}{I.M}$
Where, $\alpha $ is degree of dissociation
$D.M$ is dissociated moles
$I.M$ is initial moles
$K=\dfrac{\left[ P \right]}{\left[ R \right]}$
$K$ is the dissociation constant
$\left[ P \right]$ is the concentration of product
$\left[ R \right]$ is the concentration of reactant
Complete step by step answer:
Here, it is given that the $HI$ is dissociated $50%$
Degree of dissociation $\left( \alpha \right)=\dfrac{D.M}{I.M}$
$\alpha $ is degree of dissociation
$D.M$ is dissociated moles
$I.M$ is initial moles
Now, substituting the value we get,
$\alpha =\dfrac{50}{100}$
$\alpha =0.5$
Now, let us see the reaction
$2H{{I}_{(g)}}\rightleftharpoons {{H}_{2(g)}}+{{I}_{2(g)}}$
| $2HI$ | ${{H}_{2}}$ | ${{I}_{2}}$ | |
| Initial concentration | $2$ | $0$ | $0$ |
| Final concentration | $2-2\alpha $ | $\alpha $ | $\alpha $. |
To calculate dissociation constant $\left( K \right)$ the formula used is:
$K=\dfrac{\left[ P \right]}{\left[ R \right]}$
Where, $K$ is the dissociation constant
$\left[ P \right]$ is the concentration of product
$\left[ R \right]$ is the concentration of reactant
now, we will substitute the values in above formula we get,
At equilibrium, $K=\dfrac{{{\alpha }^{2}}}{{{\left( 2-2\alpha \right)}^{2}}}$
As we have calculated above the degree of dissociation is $0.5$
Now if we substitute the values in above formula we get,
$K=\dfrac{{{\left( 0.5 \right)}^{2}}}{{{\left( 2-2\times 0.5 \right)}^{2}}}$
On further solving,
$K=\dfrac{0.25}{{{1}^{2}}}$
$K=0.25$
So, the correct answer is “Option D”.
Additional information
The equilibrium constant is defined as the ratio of the concentration of reactants and products.
There are several characteristics of equilibrium constant:
The catalyst changes the rate of reaction but it does not change the value of equilibrium constant.
If the equilibrium is reversed then the value of equilibrium constant is reciprocal of the original equilibrium constant.
If the reaction is multiplied by the constant value $n$ then the value of equilibrium constant becomes the ${{n}^{th}}$ power of the original value.
When two or more reactions are added then the final reaction becomes the product of individual equilibrium constant.
Note: Degree of dissociation is defined as the ratio of dissociated moles to the initial moles.
Equilibrium constant is represented in many forms like equilibrium constant at constant pressure, equilibrium constant at constant concentration, equilibrium constant at constant moles.
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