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Sagot :
The equilibrium constant for this reaction Kc is 153.
Data;
- mass of hydrogen = 0.864g
- mass of iodine = 103.7g
- mass of hydrogen iodide = 92.3g
- volume of mixture = 3.51L
Number of Moles
Let us calculate the number of moles of each of the entity.
Number of moles of hydrogen
[tex]n = \frac{mass}{molar mass}\\n = \frac{0.864}{2} = 0.432 moles[/tex]
Number of moles of Iodine
[tex]n = \frac{103.7}{253.8} = 0.409 moles[/tex]
Number of moles of HI
[tex]n = \frac{92.3}{127.9} = 0.722 moles[/tex]
Equation of Reaction
[tex]H_2 + I_2 \to 2HI[/tex]
0.432 0.409 0 (Before reaction starts)
(0.432 - x) (0.409 -x) 2x (After the reaction)
[tex]2x = 0.722 moles \\x = 0.361 moles[/tex]
At Equilibrium
At equilibrium moles of H2 would be
[tex]0.432 - x = 0.432 - 0.361 = 0.071 moles[/tex]
The concentration of hydrogen would be
[tex]conc. = \frac{moles}{volume}\\conc. = \frac{0.071}{3.51}\\ conc. = 0.020M[/tex]
The concentration of iodine would be
[tex]conc. of I_2 = \frac{0.409 - 0.361}{3.51}= 0.01367M[/tex]
The concentration of hydrogen iodide would be
[tex]conc. = \frac{0.722}{3.51}\\ conc. = 0.206M[/tex]
Equilibrium Constant
This is given by
[tex]K_c = \frac{[HI]^2}{[H_2][I_2]}[/tex]
substituting the value of the concentration
[tex]K_c = \frac{[0.20]^2}{[0.020][0.01367]}\\K_c = 153[/tex]
From the calculations above, the equilibrium constant for this reaction Kc is 153.
Learn more on equilibrium constant here;
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