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Determine the standard enthlpy of the re...

Determine the standard enthlpy of the reaction `C_(3)H_(8)(g) + H_(2)(g) to C_(2)H_(6)(g) + CH_(4)(g)`. Using the given enthalpies under standard conditions.
Compound `H_(2)(g)CH_(4)(g)C_(2)H_(6)(g)C("Graphite")`
`Delta_(c)H^(0)(kJ//mol) –285.8 – 890.0 – 1560. 0 –395.5`
The standard enthalpy of formation of `C_(3)H_(8)(g)` is `–103.8 kJ//mol`

A

`-55.7 kJ`

B

`+55.7 kJ`

C

`-2060.4 kJ`

D

`+ 2060.4 kJ`

Text Solution

Verified by Experts

The correct Answer is:
A
`C_(3)H_(8)(g) + 5O_(2)(g) to 3CO_(2)(g) + 4H_(2)O(l)`
`Delta_(f)H_(H_(2)O(l)) = Delta_(C)H_(H_2(g))`
`Delta_(C)H_(C_3H_3(g)) = [3 xx Delta H_(CO_2(g)) + 4 xx Delta_(f)H_(H_(2)O(l))]`
`-[Delta_(f)H_(C_(3)H_(3)(g)) + 5 xx Delta_(f)H_(O_2(g))]`
`=[3(-393.5)+4(-285.8)]`
`-(-103.8) = -2219.9 kJ`
`Delta_(f)H_("required") = -[Delta_(C)H_(C_2H_6(g)) + Delta_(C)H_(CH_4(g))]`
`+[Delta_(C)H_(C_3H_8(g)) + Delta_(C)H_(H_2(g))]`
`= -[(-1560.0) + (-890.0)]`
`+[(-2219.9) + (-285.8)] = 55.7 kJ`.
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