Question #87759

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Answer 1 · 2016-08-29T10:01:01

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