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At a certain temperature, only 50% HI is dissociated into H2 and I2 at equilibrium. The equilibrium constant is:
- a)1.0
- b)3.0
- c)0.5
- d)0.25
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
At a certain temperature, only 50% HI is dissociated into H2 and I2 at...
The HI is dissociated with the chemical equation,
Here, a is the concentration.
The equilibrium constant of the above chemical reaction can be calculated as follows.
Here, a is the concentration.
The equilibrium constant of the above chemical reaction can be calculated as follows.
= 1/4
Thus, the value of equilibrium is 0.25.
Thus, the value of equilibrium is 0.25.
Most Upvoted Answer
At a certain temperature, only 50% HI is dissociated into H2 and I2 at...
Given:
At equilibrium, only 50% HI is dissociated into H2 and I2
To find:
Equilibrium constant
Solution:
The balanced chemical equation for the dissociation of HI is:
2HI ⇌ H2 + I2
Let the initial number of moles of HI be 'x'
At equilibrium, the number of moles of HI that dissociates into H2 and I2 = 0.5x
The number of moles of H2 and I2 formed = 0.5x/2 = 0.25x each
The expression for the equilibrium constant, Kc is:
Kc = [H2][I2]/[HI]^2
At equilibrium, the concentration of H2 and I2 is 0.25x/V (where V is the volume of the container)
The concentration of HI is (x - 0.5x)/V = 0.5x/V
Substituting the values in the expression for Kc:
Kc = (0.25x/V)^2/(0.5x/V)^2
Kc = 0.25/0.5
Kc = 0.5
Therefore, the equilibrium constant for the dissociation of HI is 0.5, which corresponds to option 'D'.
Note:
The equilibrium constant for a reaction gives us an idea about the extent of the reaction. A higher Kc value indicates that the reaction proceeds almost to completion, while a lower Kc value indicates that the reaction is not very favorable. In this case, since the value of Kc is less than 1, it suggests that the dissociation of HI is not very favorable and only about 50% of HI dissociates at equilibrium.
At equilibrium, only 50% HI is dissociated into H2 and I2
To find:
Equilibrium constant
Solution:
The balanced chemical equation for the dissociation of HI is:
2HI ⇌ H2 + I2
Let the initial number of moles of HI be 'x'
At equilibrium, the number of moles of HI that dissociates into H2 and I2 = 0.5x
The number of moles of H2 and I2 formed = 0.5x/2 = 0.25x each
The expression for the equilibrium constant, Kc is:
Kc = [H2][I2]/[HI]^2
At equilibrium, the concentration of H2 and I2 is 0.25x/V (where V is the volume of the container)
The concentration of HI is (x - 0.5x)/V = 0.5x/V
Substituting the values in the expression for Kc:
Kc = (0.25x/V)^2/(0.5x/V)^2
Kc = 0.25/0.5
Kc = 0.5
Therefore, the equilibrium constant for the dissociation of HI is 0.5, which corresponds to option 'D'.
Note:
The equilibrium constant for a reaction gives us an idea about the extent of the reaction. A higher Kc value indicates that the reaction proceeds almost to completion, while a lower Kc value indicates that the reaction is not very favorable. In this case, since the value of Kc is less than 1, it suggests that the dissociation of HI is not very favorable and only about 50% of HI dissociates at equilibrium.
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At a certain temperature, only 50% HI is dissociated into H2 and I2 at equilibrium. The equilibrium constant is:a)1.0b)3.0c)0.5d)0.25Correct answer is option 'D'. Can you explain this answer?
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