Question

Given that the equilibrium constant for the reaction
2S02(g) + O2(g) 2S03(g), has a value of
278 at a particular temperature. What is the value
of the equilibirum constant for the following reaction
at the same temperature ?
SO3(g) = SO2(g) + 2O2(g)

(1) 1.8 × 10-3 (2) 3.6 × 10-3
(3) 6.0 x 10-2 (4) 1.3 x 10-5​

Answer 1

Explanation:

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Answer 2

The value of equilibrium constant for reverse reaction is, $6.0\times 10^{-2}$ Explanation:

The given chemical equation follows:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The equilibrium constant for the above equation is 278.

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

$2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is divided by a factor of '2', the equilibrium constant of the reverse reaction will be the square root of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{c}'=\sqrt{(\frac{1}{K_c})}$

$K_{c}'=\sqrt{(\frac{1}{278})}$

$K_{c}'=0.0599\approx 6.0\times 10^{-2}$

Hence, the value of equilibrium constant for reverse reaction is, $6.0\times 10^{-2}$

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