What is the equilibrium expression for the reaction

P_{4(s)}+5O_{2(g)}\rightleftharpoons P_{4}O_{10(s)}?

  • Option 1)

    K_{c}= \frac{\left [ P_{4}O_{10} \right ]}{\left [ P_{4} \right ]\left [ O_{2} \right ]^{5}}

     

     

     

  • Option 2)

    K_{c}= \frac{\left [ P_{4}O_{10} \right ]}{5\left [ P_{4} \right ]\left [ O_{2} \right ]}

  • Option 3)

    K_{c}= \left [ O_{2} \right ]^{5}

  • Option 4)

    K_{c}=\frac{1}{ \left [ O_{2} \right ]^{5}}

 

Answers (1)

As we learnt in 

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

aA+bB\rightleftharpoons cC+dD


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

[A],\:[B],\:[C]\:[D]

are equilibrium concentration

 

 

P_{4(s)}+5O_{2(g)}\rightleftharpoons P_{4}O_{10(s)}

We know that concentration of a solid component is always taken as unity.

K_{c}= \frac{1}{\left [ O_{2} \right ]^{5}}


Option 1)

K_{c}= \frac{\left [ P_{4}O_{10} \right ]}{\left [ P_{4} \right ]\left [ O_{2} \right ]^{5}}

 

 

 

This option is incorrect

Option 2)

K_{c}= \frac{\left [ P_{4}O_{10} \right ]}{5\left [ P_{4} \right ]\left [ O_{2} \right ]}

This option is incorrect

Option 3)

K_{c}= \left [ O_{2} \right ]^{5}

This option is incorrect

Option 4)

K_{c}=\frac{1}{ \left [ O_{2} \right ]^{5}}

This option is correct

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