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

Options

A. `2C_(4)H_(10)(g) + 13O_(2)(g) rarr 3CO_(2)(g) + 10H_(2)O(l) Delta_(r) H= -2658.0 kJ mol^(-1)`
B. `C_(4)H_(10)(g) + (13)/(2) O_(2)(g) rarr 4CO_(2)(g) + 5H_(2)O (l), Delta_(r)H= -1129.9 kJ mol^(-1)`
C. `C_(4)H_(10) (g) + (13)/(2)O_(2) (g) rarr 4CO_(2)(g) + 5H_(2)O(l), Delta_(r)H= -2658.0 kJ mol^(-1)`
D. `C_(4)H_(10)(g) + (13)/(2) O_(2)(g) rarr 4CO_(2)(g) + 5H_(2)O(l), Delta_(r) H= +2658.0 kJ mol^(-1)`

Answer

C

Explanation

Given that, the complete combustion of one mole of butane is represented by thermochemical reaction as `C_(4)H_(10(g)) + (13)/(2) O_(2)(g) rarr 4CO_(2)(g) + 5H_(2)O(l)`
We have to take the combustion of one mole of `C_(4)H_(10)` and `Delta_(E)H` should be negative and have a value of 2658 kJ `mol^(-1)`.