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

`7.7*10^(-4)`

Step-by-step 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^(')=(1)/(Kc)`

2nd step:

Dividing the equation throughout by 2:

New Kc becomes:

`Kc^('')=\sqrt{Kc^(')}=(1)/(√(Kc) )`

`=\frac{1}{\sqrt{1.7*10^(6) } }=7.7*10^(-4)`

Hence the equilibrium constant is
`7.7*10^(-4)`

Answer 2

The value of equilibrium constant for reverse reaction is
`7.7* 10^(-4)`

Step-by-step 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* 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)+(1)/(2) O_2(g)`

The value of equilibrium constant for half reverse reaction is:

`K_(eq)'=((1)/(1.7* 10^6))^{(1)/(2)}=0.00077=7.7* 10^(-4)`

Hence, the value of equilibrium constant for reverse reaction is
`7.7* 10^(-4)`