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

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$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) ⟶ 4 {CO}_{2} (g) + 5 H_{2} O (l)$ ; $Δ_{c} H = + 2658 .0 kJ {mol}^{- 1}$

$2 C_{4} H_{10} (g) + 13 O_{2} (g) ⟶ 8 {CO}_{2} (g) + 10 H_{2} O (l); Δ_{c} H = - 2658.0 ke {mol}^{- 1}$

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) ⟶ 4 {CO}_{2} (g) + 5 H_{2} O (l)$ ; $Δ_{c} H = - 2658.0 kJ {mol}^{- 1}$

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) ⟶ 4 {CO}_{2} (g) + 5 H_{2} O (g)$ ; $Δ_{c} H = - 1329 .0 kJ {mol}^{- 1}$

answer is C.

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Detailed Solution

In the complete combustion of one mole of butane in the reaction given in option C, 2658 kJ of heat is released.

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) ⟶ 4 {CO}_{2} (g) + 5 H_{2} O (l)$

$Δ_{c} H = - 2658.0 kJ {mol}^{- 1}$

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