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If (1)/(2)AtoB,DeltaH=+150kJ*mol^(-1),3B...

If `(1)/(2)AtoB,DeltaH=+150kJ*mol^(-1),3B to 2C+D,DeltaH=-125kJ*mol^(-1),E+Ato2D,DeltaH=+350kJ*mol^(-1)`
then `DeltaH` of the reaction `B+D to E+2C` will be-

A

`525kJ*mol^(-1)`

B

`-175kJ*mol^(-1)`

C

`-325kJ*mol^(-1)`

D

`325kJ*mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`(1)/(2)A to B,DeltaH=+150kJ*mol^(-1)` . . [1]
`3B to 2C+D,DeltaH=-125kJ*mol^(-1)` . . [2]
`E+A to 2D,DeltaH=+350kJ*mol^(-1)` . . . [3]
Subtracting equation [3] from (2`xx`equation [1]+equation [2]), we have-
`B+D to E+2C,DeltaH=2xx150-125-350=-175kJ`.
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