The equilibrium constant for the reaction

SO_{3(g)}\rightleftharpoons SO_{2(g)}+\frac{1}{2}O_{2(g)}

is\: K_{c}= 4.9\times 10^{-2} The value of K_{c} for the reaction

2SO_{2(g)}+O_{2(g)}\rightleftharpoons 2SO_{3(g)}  will be :

  • Option 1)

    416

  • Option 2)

    2.40\times 10^{-3}

  • Option 3)

    9.8\times 10^{-2}

  • Option 4)

    4.9\times 10^{-2}

 

Answers (1)
V Vakul

As we learnt in

Equilibrium constant for the reverse reaction -

Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.

- wherein

K'_{c}=\frac{1}{K_{c}}

{K'_{c}}   is  equilibrium constant for reverse direction.

 

 

 

SO_{3(g)}\rightleftharpoons SO_{2(g)}+1/2\, O_{2(g)}

\frac{\left [ SO_{2} \right ]\left [ O_{2} \right ]^{1/2}}{\left [ SO_{3} \right ]}=K_{c}=4.9\times 10^{-2}..................(i)

SO_{2(g)}+1/2\, O_{2(g)}\rightleftharpoons SO_{3(g)}

\frac{\left [ SO_{3} \right ]}{\left [ SO_{2} \right ]\left [ O_{2} \right ]^{1/2}}=K'_{c}=\frac{1}{4.9\times 10^{-2}}..............(ii)

For\; \; 2SO_{2(g)}+O_{2(g)}\rightleftharpoons 2SO_{3(g)}

\frac{\left [ SO_{3} \right ]^{2}}{\left [ SO_{2} \right ]^{2}\left [ O_{2} \right ]}=K'^{2}_{c}=\frac{1}{4.9\times 4.9\times 10^{-4}}

=\frac{10000}{24.01}=416.49

 

 

 


Option 1)

416

This option is correct.

Option 2)

2.40\times 10^{-3}

This option is incorrect.

Option 3)

9.8\times 10^{-2}

This option is incorrect.

Option 4)

4.9\times 10^{-2}

This option is incorrect.

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