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