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C(2)H(6)(g)+3.5O(2)(g)to2CO(2)(g)+3H(2)O...

`C_(2)H_(6)(g)+3.5O_(2)(g)to2CO_(2)(g)+3H_(2)O(g)`
`DeltaS_("vap")(H_(2)O,l)=x_(1)" "cal K^(-1)` (boiling point`=T_(1)`)
`DeltaH_(f)(H_(2)O,l)" "=x_(2)`
`DeltaH_(f)(CO_(2))" "=x_(3)`
`DeltaH_(f)(C_(2)H_(6))" "=x_(4)`
Hence, `DeltaH` for the reaction is :

A

`2x_(3)+3x_(2) -x_(4)`

B

`2x_(3)+3x_(2)-x_(4)+3x_(1)T_(1)`

C

`2x_(3) +3x_(2)-x_(4) -3x_(1)T_(1)`

D

`x_(1)T_(1) + x_(2) +x_(3) +x_(4)`

Text Solution

Verified by Experts

The correct Answer is:
B
From `Delta S_("vap"), Delta H = Delta S xx T=3X_(1) T_(1)("For "3H_(2)O_((g))) and Delta H_("total")=2X_(3)+3X_(2)-X_(4)+3X_(1)T_(1)`
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