Question

Calculate `P_{SO_2}` at equilibrium at 298 K?

For the following reaction run at 298 K; a mixture of equilibrium contains `P_{O_2}=0.50` atm, `P_{SO_2}=2.0` atm. Calculate `P_{SO_2}` at equilibrium at 298 K?
`2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)`
`\Delta G°_{f,SO_3(g)}=-371\frac{kJ}{mol}`
`\Delta G°_{f,SO_2(g)}=-300\frac{kJ}{mol}`

Answer 1

Well, apparently all the `"SO"_2` was consumed...

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First, write the reaction and equilibrium constant:

`2"SO"_2(g) + "O"_2(g) rightleftharpoons 2"SO"_3(g)`

`K_P = P_(SO_3)^2/(P_(SO_2)^2P_(O_2))`

What we want to solve for is the (always positive) partial pressure of `"SO"_2` (not `"SO"_3`, apparently):

`P_(SO_2) = sqrt(P_(SO_3)^2/(K_PP_(O_2))`

I assume you gave `P_(O_2)` and `ulul(P_(SO_3))`...

We know all of these quantities except for `K_P`. Remember that gas-phase reactions have `K -= K_P`, so:

`DeltaG_(rxn)^@("gas-phase") = -RTlnK_P`

We already have enough information to find `DeltaG^@`, so we can find `K_P` afterwards. These `DeltaG_f^@` are correct and the reaction is balanced.

`DeltaG_(rxn)^@ = ["2 SO"_3(g) cdot DeltaG_f("SO"_3(g))^@] - ["2 SO"_2(g) cdot DeltaG_f("SO"_2(g))^@ + "1 O"_2(g) cdot "0 kJ/mol O"_2(g)]`

`= [-"742 kJ/mol"] - [-"600 kJ/mol"]`

`= -"142 kJ/mol"`

Therefore,

`K_P = e^(-DeltaG_(rxn)^@//RT)`

`= e^(-(-"142 kJ/mol")//("0.008314 kJ/mol"cdot"K" cdot "298 K")`

`= e^(57.3)`

As a result, we get the nearly zero partial pressure of `"SO"_2`:

`P_(SO_2) = sqrt(("2.0 atm")^2/(e^(57.3)("0.50 atm"))`

`=` `ul(1.02 xx 10^(-12) "atm")`

i.e. basically zero...