contestada

Given the equilibrium constants for the following reactions: 4Cu(s) + O2(g) <--> 2Cu2O(s), K1 2CuO(s) <--> Cu2O(s) + 1/2 O2(g), K2 what is K for the system 2Cu(s) + O2(g) <--> 2CuO(s) equivalent to?

Respuesta :

Answer:

[tex]K = 1/K_{1}\times 2K_{2} = 2K_{2}/K_{1}[/tex]

Explanation:

                   [tex]4Cu{(s)} + O_{2}(gas) \leftrightarrow 2Cu_{2}O_{(s)}[/tex]   [tex],K_{1}[/tex] assuming equation (1)

                   [tex]2CuO_{(s)} \leftrightarrow Cu_{2}O{(s)} + 1/2 O_{2}(gas) ,K_{2}[/tex]  assuming equation (2)

upon reverting equation (1) and multiplying equation (2) by 2

             

             [tex]2Cu_{2}O_{(s)} \leftrightarrow 4Cu{(s)} + O_{2}(gas)[/tex]        ,[tex],1/K_{1}[/tex] assuming equation (3)

           [tex]4CuO_{(s)} \leftrightarrow 2Cu_{2}O{(s)} + O_{2}(gas)[/tex]     [tex],2K_{2}[/tex]   assuming equation (4)

upon adding equation (3) and (4) , give  equation (5)

           [tex]4CuO_{(s)} \leftrightarrow 4Cu{(s)} + 2O_{2}(gas)[/tex]  [tex]K_{3}=K/2[/tex]  , equation (5)

  divide equation (5) by 2

           [tex]2CuO_{(s)} \leftrightarrow 2Cu{(s)} + O_{2}(gas)[/tex]   [tex]K = 1/K_{1}\times 2K_{2}[/tex]

                      [tex]K = 1/K_{1}\times 2K_{2} = 2K_{2}/K_{1}[/tex]

                     

adioabiola

The equilibrium constant of the target reaction is equivalent to: K = K1/2K2

According to the question;

  • The target equation is;

2Cu(s) + O2(g) <--> 2CuO(s)

Given;

  • 4Cu(s) + O2(g) <--> 2Cu2O(s), K1.........eqn(1)

  • 4Cu(s) + O2(g) <--> 2Cu2O(s), K1.........eqn(1)2CuO(s) <--> Cu2O(s) + 1/2 O2(g), K2......eqn(2)

By observation; to arrive at the target equation;

We must reverse equation (2) and divide equation 1 by 2 so that we have;

  • 2Cu(s) + 1/2O2(g) <--> Cu2O(s),
  • 2Cu(s) + 1/2O2(g) <--> Cu2O(s), k1/2.........eqn(3)

  • 2Cu(s) + 1/2O2(g) <--> Cu2O(s), k1/2.........eqn(3)Cu2O(s) + 1/2 O2(g) <--> 2CuO(s),
  • 2Cu(s) + 1/2O2(g) <--> Cu2O(s), k1/2.........eqn(3)Cu2O(s) + 1/2 O2(g) <--> 2CuO(s), 1/K2.....eq4

The aggregate of eqn(3) and (4) therefore becomes;

  • 2Cu(s) + 1/2O2(g) + Cu2O(s) + 1/2 O2(g) <--> Cu2O(s) + 2CuO(s)

By eliminating the spectator compound; Cu2O(s); we have;

  • 2Cu(s) + O2(g) <--> 2CuO(s)

2Cu(s) + O2(g) <--> 2CuO(s)where; K = k1/2 × 1/K2

2Cu(s) + O2(g) <--> 2CuO(s)where; K = k1/2 × 1/K2K = k1/2k2

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