Question

Question #87759

Answer 1

Trying to present an answer assuming slight change in the given reaction to keep consistency with the unit of equilibrium constant.

The reaction and ICE table

`" "A" "+" "2 B" " rightleftharpoons" "C`

I`" "x" mol "+" "4""mol " 0 " mol"`

C`""-1" mol "+" "-2""mol " +1 " mol"`

E`""(x-1)" mol "+" "2""mol " 1 " mol"`

Where x mole is the amount of A to be mixed with 4 moles of B initially

The volume of the reaction mixture given is `V=5dm^3`.

So at equilibrium the concentration of the components will be

`[A]=(x-1)/V"mol"/"dm"^3`

`[B]=2/V"mol"/"dm"^3`

`[C]=1/V"mol"/"dm"^3`

Now equilibrium constant

`K_c=([C])/([A][B]^2)=(1/V)/((x-1)/V*(2/V)^2)`

`=>K_c=V^2/(4(x-1))dm^6mol^-2`

Inserting the value of `K_c and V` we get

`0.25=5^2/(4(x-1))`

`=>x-1=25=>x=26`

So the amount of A to be added to 4 moles of B is 26 moles