Question

The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant.2 SO2(g) + O2(g) ↔ 2 SO3(g) Kc = 1.7 × 106SO3(g) ↔ 1/2 O2(g) + SO2(g) Kc = ?1.3 × 10^31.2 × 10^-68.57.7 × 10^-43.4 × 10^2

Answer 1

Answer:

The value of equilibrium constant for reverse reaction is $7.7\times 10^{-4}$

Explanation:

The given chemical equation follows:

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

The equilibrium constant for the above equation is $1.7\times 10^6$.

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 multiplied by a factor of '1/2', the equilibrium constant of the reverse reaction will be the square root of the equilibrium constant  of initial reaction.

So,

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

The value of equilibrium constant for half reverse reaction is:

$K_{eq}'=(\frac{1}{1.7\times 10^6})^{\frac{1}{2}}=0.00077=7.7\times 10^{-4}$

Hence, the value of equilibrium constant for reverse reaction is $7.7\times 10^{-4}$

Answer 2

Answer:

$7.7\times10^{-4}$

Explanation:

The equation for which we have to find Kc is obtained by two - step transformation of the equation whose Kc is given.

1st step:

Reversing the reaction:

By reversing the reaction the reactants become products and vice-versa.

The new equilibrium constant will be:

$Kc^{'}=\frac{1}{Kc}$

2nd step:

Dividing the equation throughout by 2:

New Kc becomes:

$Kc^{''}=\sqrt{Kc^{'}}=\frac{1}{\sqrt{Kc} }$

$=\frac{1}{\sqrt{1.7\times10^{6} } }=7.7\times10^{-4}$

Hence the equilibrium constant is $7.7\times10^{-4}$