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An absent-mined prodessie, Mr Waage, took elements A and B in a reaction vessel at room temperature, to study the reaction`A+ 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 ? |
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Answer» `A + 2B hArr 2C +D` Initial concentration a 1.5a 0 0 Equilibrium concentration a-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` |
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