<span>0.0750 M Na3PO4 as this solution would contain 3 Na+ and 1 PO4- ions per mole of Na3PO4 for an effective total ion concentration of 4 x .0750 or .300 M. The K2SO4 has three total ions or a concentration of .300 M as well. Hope it helps. </span>
<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
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