Question

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

• A. $2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(l) \Delta_c H = -2658.0 \, kJ \, mol^{-1}$
• B. $C_4H_{10}(g) + \frac{13}{2} O_2(g) \rightarrow 4CO_2 (g) + 5H_2 O(g) \Delta_c H =- 1329.0 kJ \, mol^{-1}$
• C. $C_4H_{10}(g) + \frac{13}{2} O_2(g) -\rightarrow 4CO_2(g) + 5H_2 O(l) \, \Delta_c H = - 2658.0 \, kJ \, mol^{-1}$
• D. $C_4H_{10} (g) + \frac{13}{2} O_2 (g) \rightarrow 4CO_2 (g) + 5H_2O(l) \Delta_c H = + 2658.0 \, kJ \, mol^{-1}$

Answer 1

Answer: (C)

$C _{4} H _{10}( g )+\frac{13}{2} O _{2( g )} \longrightarrow 4 CO _{2( g )}+5 H _{2} O _{(1)} ;$
$\Delta_{c} H =-2658.0\, kJ / mol$
The complete combustion of 1 mole of butane is represented by
$C _{4} H _{10( g )}+\frac{13}{2} O _{2( g )} \longrightarrow 4 CO _{2( g )}+5 H _{2} O _{(1)} ;$
$\Delta_{ c } H =-2658.0\, kJ / mol$
$\Delta_{ c } H$ should be negative and have a value of $2658\, kJ / mol$.