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A 6.0-l vessel was found to contain 1.0 mol brcl3, 2.0 mol br2 and 6.0 mol cl2. what is the equilibrium constant, kc, for this equilibrium mixture for the reaction 2brcl3(g) ? br2(g) + 3cl2(g)?

Respuesta :

kenmyna

The equilibrium constant (Kc) for this equilibrium mixture for the reaction

2 BrCl3 ⇔ Br2 + 3 Cl2 is 11.94


calculation


Kc=( concentration of the product raised to their coefficients/ concentration of the reactant raised to their coefficient)


Kc={( Br2) Cl2)^2} /(BrCl3)^2


find the concentration of the reactant and products

=moles/volume in liters


BrCl3 = 1.0 /6.0 = 0.167M

Br= 2.0 /6.0 =0.333 M

Cl2=6.0/6.0 =1.00 M


Kc is therefore = {(0.333) (1.00)^3 / (0.167)^2} = 11.94

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