<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61
<u>Explanation:</u>
We are given:
Initial moles of iodine gas = 0.100 moles
Initial moles of hydrogen gas = 0.100 moles
Volume of container = 1.00 L
Molarity of the solution is calculated by the equation:



Equilibrium concentration of iodine gas = 0.0210 M
The chemical equation for the reaction of iodine gas and hydrogen gas follows:

<u>Initial:</u> 0.1 0.1
<u>At eqllm:</u> 0.1-x 0.1-x 2x
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
![[HI]_{eq}=2x=(2\times 0.079)=0.158M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.079%29%3D0.158M)
![[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D%280.1-x%29%3D%280.1-0.079%29%3D0.0210M)
![[I_2]_{eq}=0.0210M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.0210M)
Putting values in above expression, we get:

Hence, the value of equilibrium constant for the given reaction is 56.61
Answer:
Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state
Answer:
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