Respuesta :
Answer: The moles of [tex]CO_2[/tex] added to the system is 7.13 moles
Explanation:
We are given:
Moles of [tex]CO_2[/tex] at equilibrium = 1.00 moles
Moles of [tex]H_2[/tex] at equilibrium = 1.00 moles
Moles of [tex]H_2O[/tex] at equilibrium = 2.40 moles
Moles of [tex]CO[/tex] at equilibrium = 2.40 moles
Volume of the container = 4.00 L
Concentration is written as:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}[/tex]
The given chemical equation follows:
[tex]CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}[/tex]
Putting values in above equation, we get:
[tex]K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76[/tex]
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Molarity of CO = 0.791 mol/L
Volume of solution = 4.00 L
Putting values in above equation, we get:
[tex]0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol[/tex]
Extra moles of CO = (3.164 - 2.40) = 0.764 moles
Let the moles of [tex]CO_2[/tex] needed be 'x' moles.
Now, equilibrium gets re-established:
[tex]CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)[/tex]
Initial: 1.00 1.00 2.40 2.40
At eqllm: (0.236+x) 0.236 3.164 3.164
Again, putting the values in the expression of [tex]K_c[/tex], we get:
[tex]5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13[/tex]
Hence, the moles of [tex]CO_2[/tex] added to the system is 7.13 moles
7.13 moles of CO₂ must be added to the system to bring the equilibrium CO concentration to 0.791 mol/L.
How we calculate moles from concentration?
If the concentration of any substance is present in the form of molarity (M) then their moles (n) will be calculated as:
M = n/V, where
V = volume
Given chemical reaction is and moles of their participants are:
CO₂(g) + H₂(g) ⇄ H₂O(g) + CO(g)
Moles of CO₂ = 1 mole
Moles of H₂ = 1 mole
Moles of H₂O = 2.40 moles
Moles of CO = 2.40 moles
Volume of reaction vessel = 4L
Equilibrium constant of the given reaction will be calculated as the ration of the product of the concentration of products to the product of the concentration of reactants and concentration will be written by using the above formula as:
Kc = (2.40/4)×(2.40/4) / (1/4)(1/4) = 5.76
When equilibrium gets re established then,
Given, molarity of CO = 0.791 mol/L
Then, moles of C0 = 0.791mol/L × 4L = 3.164mol
CO₂(g) + H₂(g) ⇄ H₂O(g) + CO(g)
Initial: 1 1 2.40 2.40
Equilibrium: 0.236+x 0.236 3.164 3.164
Extra moles of CO = 3.164 - 2.40 = 0.764 moles
Let, required moles of needed CO₂ be x.
Now we calculate the value of x, by putting all values again for the equilibrium constant as:
Kc = (3.164/4)×(3.164/4) / (0.236+x/4)×(0.236/4)
5.76 = 10.011 / 0.056+0.236x
x = 7.13 moles
Hence, needed moles of CO₂ is 7.13 moles.
To know more about molarity, visit the below link:
https://brainly.com/question/16343005
