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Given the equilibrium constants for the following two reactions at a 298K:NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) Kc=40NiO(s) +CO(g) ⇌ Ni(s) +CO2(g) Kc=600Calculate the value for the equilibrium constant, Kc, for the reaction:CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

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Answer:

The value  is  [tex]K_C = \frac{40}{600}[/tex]

Explanation:

From the question

   The equation given is  

            NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) Kc=40          (1)

            NiO(s) +CO(g) ⇌ Ni(s) +CO2(g) Kc=600         (2)

Generally the reverse of the second equation as shown below

            Ni(s) +CO2(g) ⇌  NiO(s) + CO(g)                 (3)

The equilibrium constant becomes    [tex]K_c ' = \frac{1}{600}[/tex]  

 Now  adding  1  and  3 we obtain

      NiO(s) + H2(g)+ Ni(s) +CO2(g) ⇌ Ni(s) + H2O(g)+NiO(s) + CO(g)        

               CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

Hence the equilibrium constant for the resulting equation is mathematically evaluated as

           [tex]K_C = K_c' * K_c[/tex]

=>       [tex]K_C = \frac{1}{600} * 40[/tex]

=>         [tex]K_C = \frac{40}{600}[/tex]

       

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