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A reaction mixture in a 3.51 l flask at a certain temperature initially contains a mixture of 0.864 grams of h2 and 103.7 grams of i2. at equilibrium, the flask contains 92.3 grams if hi. calculate the equilibrium constant for this reaction at this temperature. show your work.

Respuesta :

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.

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