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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)

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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