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C(2)H(6)(g) + 3.5O(2)(g) rarr 2CO(2)(g) ...

`C_(2)H_(6)(g) + 3.5O_(2)(g) rarr 2CO_(2)(g) + 3H_(2)O(g)`
`DeltaS_("vap") (H_(2)O,l) = "x"_(1)calK^(-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

`2"x"_(3) + 3"x"_(2)-"x"_(4)`

B

`2"x"_(3) + 3"x"_(2)-"x"_(4) + 3"x"_(1)T_(1)`

C

`2"x"_(3) + 3"x"_(2)-"x"_(4) - 3"x"_(1)T_(1)`

D

`"x"_(1)T_(1) + "X"_(2) + "X"_(3) - "x"_(4)`

Text Solution

Verified by Experts

The correct Answer is:
B
`H_(2)O(I) rarr H_(2)O(g) DeltaH_("vap") = DeltaS_("vap") T_(B.P) = "x"_(1)T_(1)`
`DeltaH_(f)(H_(2)O,g) = DeltaH_(f)(H_(2)O,I) + DeltaH_("vap") = "x"_(2) + "x"_(1)"T"_(1)`
`DeltaH_("reaction") = 2DeltaH_(f) (CO_(2),g) + 3DeltaH_(f)(H_(2)O,g) - DeltaH_(f)(C_(2)H_(6),g)`
`" "=2"x"_(3) + 3("x"_(2) + "x"_(1)"T"_(1)) - "x"_(4)=2"x"_(3) + 3"x"_(2) + 3"x"_(1)"T"_(1) - "x"_(4).`
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