Answer is: the equilibrium constant is 57,14.
Chemical reaction: Br₂ + H₂ → 2HBr.
c₀(Br₂) = 70,31 g ÷ 160 g/mol ÷ 2 L.
c₀(Br₂) = 0,22 mol/L.
c₀(H₂) = 1,374 g ÷ 2 g/mol ÷ 2 L.
c₀(H₂) = 0,34 mol/L.
c(H₂) = 0,566 g ÷ 2 g/mol ÷ 2 L = 0,14 mol/L, so 0,2 mol/L reacts.
c(HBr) = 2 · (0,34 mol/L - 0,14 mol/L) = 0,40 mol/L; from chemical reaction n(H₂) : n(HBr) = 1 : 2.
c(Br₂) = 0,22 mol/L - 0,2 mol/L = 0,02 mol/L. n(Br₂) : n(H₂) = 1 : 1.
Kc = c(HBr)² ÷ (c(Br₂) · c(H₂)).
Kc = (0,4 mol/L)² ÷ (0,02 mol/L) · (0,14 mol/L).
Kc = 57,14.
The equilibrium constant or Kc is 57.14.
Kc is the ratio of equilibrium product concentrations to equilibrium reactant concentrations.
The reaction is
[tex]\rm Br_2 + H_2 = 2HBr.[/tex]
[tex]\rm c^\circ(Br_2) = \dfrac{70.31 g/160 g/mol}{2L} = 0.22 mol/L[/tex]
[tex]\rm c^\circ(H_2) = \dfrac{0.566 g g/2 g/mol}{2L} = 0.14 mol/L[/tex]
[tex]\rm c(HBr) = 2 \times (0.34 mol/L - 0.14 mol/L) = 0.40 mol/L\; from\; chemical\; reaction n(H_2) : n(HBr) = 1 : 2.[/tex]
[tex]\rm c(Br_2) = 0.22 mol/L - 0.2 mol/L = 0.02 mol/L \times n(Br_2) : n(H_2) = 1 : 1.[/tex]
Calculating the Kc
[tex]\rm Kc = \dfrac{c(HBr)^2}{ c(Br_2) } \times c(H_2)\\\\\\\rm Kc = \dfrac{(0.4 mol/L)^2}{(0,02 mol/L)} \times (0.14 mol/L)\\\\Kc = 57.14[/tex].
Thus, the kc is 57.14
Learn more about Kc here
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