An absent-mined prodessie, Mr Waage, took elements A and B in areaction vessel at room temperature, to study the reactionA+ 2B rArr 2C +D. He took the concentration of B as 1.5 times the concentration of A. After the reaction reached equilbrium, he round that the concentrations of A and D were equal. However, he forgot to calculate K_(c) and removed oneof the products from the mixture. Now, can you calculate K_(c) for the equilibrium attained in his experiment and help him out ?

An absent-mined prodessie, Mr Waage, took elements A and B in areactio

1.	An absent-mined prodessie, Mr Waage, took elements A and B in areaction vessel at room temperature, to study the reactionA+ 2B rArr 2C +D. He took the concentration of B as 1.5 times the concentration of A. After the reaction reached equilbrium, he round that the concentrations of A and D were equal. However, he forgot to calculate K_(c) and removed oneof the products from the mixture. Now, can you calculate K_(c) for the equilibrium attained in his experiment and help him out ?
Answer» Solution :`A + 2B hArr2C +D` Initialconcentrationa 1.5a 0 0 EQUILIBRIUM concentrationa-x 1.5a-2x 2x x At equilibrium, the concentration of A and that of D are equal `therefore a-x = x IMPLIES a=2x ` or ` x=a//2` `K_(c)=((2x)^(2)x)/((a-x)(15a-2x)^(2))` as a=2x `implies (4x^(2).x)/(x xx (x)^(2))implies K_(c) = 4`