Question

For the following reaction at equilibrium SO3(g) + NO(g) = SO2(g) + NO2(g)It is found that [SO2] = 0.70 M and [NO] = 1.20 M. Calculate the equilibrium constant for the readction knowing that the initial concentration were [SO3] = 2.55 M and [NO] = 1.90 M.

Answer 1

`K=0.14`

Step-by-step explanation:

Hello!

In this case, for the undergoing chemical reaction, we can write the equilibrium expression via:

`K=([SO_2][NO_2])/([SO_3][NO])`

Whereas the equilibrium concentration of both SO3 and NO are 2.55 M and 1.90 M respectively, it means that the extent of reaction
`x` is:

`x=1.90M-1.20M=0.7M`

Because the equilibrium expression in terms of the reaction extent is:

`K=(x*x)/(([SO_3]_0-x)([NO]_0-x))`

It means that the concentration of SO3, NO, SO2 and NO2 at equilibrium are:

`[SO_3]=2.55M-0.70M=1.85M`

`[NO]=1.20M`

`[SO_2]=0.70M`

`[NO_2]=0.70M`

Thus, the equilibrium constant for such reaction is:

`K=(0.70*0.70)/(1.85*1.90)\\\\K=0.14`

Best regards!