Respuesta :
Answer: The equilibrium moles of HI is 2.168 moles
Explanation:
We are given:
Initial moles of hydrogen gas = 1.39 moles
Initial moles of iodine = 1.39 moles
Molarity is calculated by using the formula:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Volume of the container = 5.00 L
The given chemical equation follows:
[tex]H_2+I_2\rightleftharpoons 2HI[/tex]
Initial: 1.39 1.39
At eqllm: 1.39-x 1.39-x 2x
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[HI]^2}{[I_2][H_2]}[/tex]
We are given:
[tex]K_c=50.2[/tex]
[tex][HI]_{eq}=\frac{2x}{5.00}[/tex]
[tex][H_2]_{eq}=\frac{(1.39-x)}{5.00}[/tex]
[tex][I_2]_{eq}=\frac{(1.39-x)}{5.00}[/tex]
Putting values in above expression, we get:
[tex]50.2=\frac{(2x/5)^2}{((1.39-x)/5)\times ((1.39-x)/5)}\\\\x=1.084,1.94[/tex]
Neglecting the value of x = 1.94 because equilibrium concentration cannot be greater than initial concentration
So, equilibrium moles of [tex]HI=2x=(2\times 1.084)=2.168moles[/tex]
Hence, the equilibrium moles of HI is 2.168 moles
The number of moles is the ratio of the mass and the molar mass of the substance. The equilibrium mole number of HI is 2.168 moles.
What is the equilibrium constant (Kc)?
The equilibrium constant (Kc) is the rate of the reaction respective to the concentration of the reactants and the products formed.
The chemical reaction can be shown as:
[tex]\rm H_{2} + I_{2}\leftrightharpoons 2HI[/tex]
Given,
The equilibrium constant (Kc) = 50.2
Initial moles of [tex]\rm H_{2}[/tex] = 1.39 moles
At equilibrium, the moles of hydrogen gas = 1.39 - x
Initial moles of [tex]\rm I_{2}[/tex] = 1.39 moles
At equilibrium, the moles of iodine = 1.39 - x
At equilibrium, the moles of hydrogen iodide = 2x
The equilibrium constant for the chemical reaction can be shown as:
[tex]\rm K_{c}= \dfrac{[HI]^{2}}{[I_{2}][H_{2}]}[/tex]
We know that,
[tex]\rm [HI]_{eq} = \dfrac{2x}{5}[/tex]
[tex]\rm [H_{2}]_{eq} = \dfrac{1.39-x}{5}[/tex]
[tex]\rm [I_{2}]_{eq} = \dfrac{1.39-x}{5}[/tex]
Substituting values in the above we get:
[tex]\begin{aligned}50.2 &= \dfrac{(\dfrac{2x}{5})^{2}}{(\dfrac{1.39-x}{5})(\dfrac{1.39-x}{5})}\\\\\rm x &= 1.084\end{aligned}[/tex]
Therefore, the equilibrium moles of hydrogen iodide is 2.168 moles.
Learn more about the equilibrium constant here:
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