The reaction `A to B` , `DeltaH=+24 kJ//"mole"`. For the reaction `B to C`, `DeltaH=-18 kJ//"mole"`. The decreasing order of enthalpy of `A`, `B`, `C` follow the order

Options

A. `A`, `B`, `C`
B. `B`, `C`, `A`
C. `C`, `B`, `A`
D. `C`, `A`, `B`

Answer

B

Explanation

Let `Delta_(f)H(A)=x kJ mol^(-1)`
`A to B`, `DeltaH=+24 kJ mol^(-1)`
`DeltaH=Delta_(f)H(B)-Delta_(f)H(A)`
`Delta_(f)H(B)=DeltaH+Delta_(f)H(A) `
`:. Delta_(f)H(B)=(x+24)kJ mol^(-1)`
` B to C`, `DeltaH=-18 kJ mol^(-1)`
`DeltaH=Delta_(f)H(C)-Delta_(f)H(B)`
`Delta_(f)H(C )=DeltaH+DeltaH_(f)H(B)`
`=[-18+(x+24)]kJ mol^(-1)`
`=(x+6)kJ mol^(-1)`
`:.` Decreasing order ,
`B gt C gt A`