1.

Equilibrium concentrations of A, B and C in a reversible reaction `3A+BhArr 2C+D` are `0.03`, `0.01`, and `0.008 mol L^(-1)`. Calculate the initial concentration of A?A. `0.014`B. `0.042`C. `0.084`D. `0.343`

Answer» Correct Answer - B
`{:(,3A+BhArr2C+D),("Initial (M)"," a b 0 0"),("Change (M)"," -3x -x +2x +x"),("Equilibrium (M)",bar("(a-3x) (b-x) 2x x ")):}`
According to the data,
`2x=0.008`
`x=0.004`
`:. (a-3x)=[a-3(0.004)]=0.03`
`a=0.03+0.012`
`=0.042`
Alternative method: According to the stoichiometry of the reaction, 2 mol C consume 3 mol A.
1 mol C consumes `3//2 mol A`
`0.008 mol C` consume `3/2xx(0.008)=0.012 mol A`
`:.` Initial concentration of `A`
`=` Equilibrium concentration `+` Concentration reacted at equilibrium
`=0.03+0.012=0.042`


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