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During complete combustion of one mole o...

During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is

A

`2C_(4) H_(10(g)) + 13O_(2(g)) to 8CO_(2(g)) + 10H_(2) O_((l)) , Delta_(c) H= - 2658.0 "kJ mol"^(-1)`

B

`C_(4) H_(10(g)) + (13)/(2)O_(2(g)) to 4CO_(2(g)) + 5H_(2) O_((l)) , Delta_(c) H= - 1329.0 "kJ mol"^(-1)`

C

`C_(4) H_(10(g)) + (13)/(2)O_(2(g)) to 4CO_(2(g)) + 5H_(2) O_((l)) , Delta_(c) H= - 2658.0 "kJ mol"^(-1)`

D

`C_(4) H_(10(g)) + (13)/(2)O_(2(g)) to 4CO_(2(g)) + 5H_(2) O_((l)) , Delta_(c) H= + 2658.0 "kJ mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
Given that, the complete combustion reaction of one mole of butane is represented thermodynamically as :
`C_(4) H_(10(g)) + (13)/(2) O_(2(g)) to 4CO_(2(g)) to 4CO_(2(g)) + 5H_(2) O_((l))`
We have to take the combustion of one mole of `C_(4) H_(10) and Delta_( C) H` should be negative and have a value of 2658 `"kJ mol"^(-1)`
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