Question:
During complete combustion of one mole of butane, 2658 kJ of heat is released.
The thermochemical reaction for above change is
(i) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l) ∆cH = –2658.0 kJ mol–1
(ii) C4H10(g) +13/2 O2 (g) → 4CO2 (g) + 5H2O (g) ∆cH = –1329.0 kJ mol–1
(iii) C4H10(g) +13/2 O2 (g) → 4CO2 (g) + 5H2O (l) ∆cH = –2658.0 kJ mol–1
(iv) C4H10 (g) +13/2 O2 (g) → 4CO2 (g) + 5H2O (l) ∆cH = +2658.0 kJ mol–1
Solution:
Option (iii) C4H10(g) +13/2 O2 (g) → 4CO2 (g) + 5H2O (l) ∆cH = –2658.0 kJ mol–1 is the answer.
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