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bogdanovich [222]
3 years ago
11

The reaction: 2 SO2(g) + O2(g) --> 2 SO3(g) has an equilibrium constant of K1. What is the K value for the reaction: SO3(g) -

-> SO2(g) + ½ O2(g)?
K1^½
1/K1
½ K1
(1/K1)^½
Chemistry
1 answer:
Murljashka [212]3 years ago
8 0

<u>Answer:</u> The value of equilibrium constant for reverse reaction is (\frac{1}{K_1})^{1/2}

<u>Explanation:</u>

The given chemical equation follows:

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

The equilibrium constant for the above equation is K_1

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

SO_3(g)\rightarrow SO_2(g)+\frac{1}{2}O_2(g)

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

If the equation is multiplied by a factor of '\frac{1}{2}', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant  of initial reaction.

The value of equilibrium constant for reverse reaction is:

K_{eq}'=(\frac{1}{K_1})^{1/2}

Hence, the value of equilibrium constant for reverse reaction is (\frac{1}{K_1})^{1/2}

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