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zepelin [54]
2 years ago
12

Suppose that in an equilibrium mixture of HCl, Cl2, and H2, the concentration of H2 is 1.0 x 10-11 mol-L-1and that of Cl2 is 2.0

x 10-10 mol-L-1. What is the equilibrium molar concentration of HCl at 500 K, given Kc = 4.0 x 1018 for H2(g) +Cl2(g) ⇆ 2HCl(g).
Chemistry
1 answer:
bekas [8.4K]2 years ago
3 0

Considering the definition of Kc, the equilibrium molar concentration of HCl at 500 K is 0.0894 \frac{mol}{L}.

The balanced reaction is:

H₂(g) +Cl₂(g) ⇆ 2 HCl(g)

Equilibrium is a state of a reactant system in which no changes are observed as time passes, despite the fact that the substances present continue to react with each other. In other words, reactants become products and products become reactants and they do so at the same rate.

In other words, chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed.

The concentration of reactants and products at equilibrium is related by the equilibrium constant Kc. Its value in a chemical reaction depends on the temperature and the expression of a generic reaction aA + bB ⇄ cC is

K_{c} =\frac{[C]^{c} x[D]^{d} }{[A]^{a} x[B]^{b} }

That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case, the constant Kc can be expressed as:

K_{c} =\frac{[HCl]^{2} }{[H_{2} ]x[Cl_{2} ] }

You know that in an equilibrium mixture of HCl, Cl₂, and H₂:

  • the concentration of H₂ is 1.0×10⁻¹¹ \frac{mol}{L}
  • the concentration of Cl₂ is 2.0×10⁻¹⁰ \frac{mol}{L}
  • Kc=4×10¹⁸

Replacing in the expression for Kc:

4x10^{18}  =\frac{[HCl]^{2} }{[1x10^{-11}  ]x[2x10^{-10} ] }

Solving:

4x10^{18}  =\frac{[HCl]^{2} }{2x10^{-21}  }

4x10^{18} x 2x10^{-21}=[HCl]^{2}

8x10^{-3} =[HCl]^{2}

\sqrt[2]{8x10^{-3}}  =[HCl]

0.0894 \frac{mol}{L}= [HCl]

Finally, the equilibrium molar concentration of HCl at 500 K is 0.0894 \frac{mol}{L}.

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