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Given that the equilibrium constant for the reaction

2SO2(g)+O2(g) \rightleftharpoons 2 SO3(g)

has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature?

SO3(g) \rightleftharpoons SO2(g)+1/2 O2(g)

  • Option 1)

    3.6 x 10-3

  • Option 2)

    6.0 x 10-2

  • Option 3)

    1.3 x 10-5

  • Option 4)

    1.8 x 10-3

 

Answers (1)

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_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3}(g),\ \, \, \, \, \, \, K=278

\therefore    for  SO_{3}(g)\rightleftharpoons SO_{2}(g)+\frac{1}{2}O_{2}(g),\ \: \: \: \: \: K'=\left ( \frac{1}{K} \right )^{1/2}

K'=\left ( \frac{1}{278} \right )^{1/2}=6 \times 10^{-2}


Option 1)

3.6 x 10-3

Incorrect

Option 2)

6.0 x 10-2

Correct

Option 3)

1.3 x 10-5

Incorrect

Option 4)

1.8 x 10-3

Incorrect

Posted by

Vakul

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