Question

Question 10 0 / 3.5 points Many high temperature studies have been carried out on the equilibrium of the reaction: 2SO2(g) + O2(g) = 2 SO3(g) In one study the reaction vessel initially contained (5.000x10^-3) M SO2, (2.50x10^-3) MO2, and no SO3. If it was determined that at equilibrium the SO2 concentration was (2.8x10^-3) M, determine Kc at this temperature for the reaction as written. • Answers must be written in scientific notation • Write your answer using ONE decimal place (TWO significant figures), even if this is not the correct number of significant figures (e.g., 3.4E-6 or 3.4 x 10-6). • Do NOT use spaces. • Do NOT include units.

Answer 1

Answer:
`4.4* 10^(2)`

Step-by-step explanation:

The chemical reaction follows the equation:

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

t = 0
`5.000* 10^(-3)`
`2.50* 10^(-3)` 0

At eqm
`(5.000* 10^(-3)-2x)`
`(2.50* 10^(-3)-x)` (2x)

The expression for
`K_c` for the given reaction follows:

`K_c=([SO_3]^2)/([SO_2]^2[O_2])`

`K_c=([2x]^2)/([5.000* 10^(-3)-2x]^2[2.50* 10^(-3)-x])`

Given :
`[SO_2]_(eqm)=2.80* 10^(-3)`

`5.000* 10^(-3)-2x=2.80* 10^(-3)`

`x=1.1* 10^(-3)`

Putting the values we get:

`K_c=([2* 1.1* 10^(-3)]^2)/([5.000* 10^(-3)-2* 1.1* 10^(-3)]^2[2.50* 10^(-3)-1.1* 10^(-3)])`

`K_c=4.4* 10^2`

Therefore, the equilibrium concentration
`4.4* 10^(2)`