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Equilibrium concentrations of A, B and C...

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`

Text Solution

Verified by Experts

The correct Answer is:
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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