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

Question

During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is (A) 2C4H10(g) + 13O2(g) arrow 8CO2(g) + 10H2O(l) Δc H = -2658.0 kJ mol-1 (B) C4H10(g) + (13/2) O2(g) arrow 4CO2 (g) + 5H2 O(g) Δc H =- 1329.0 kJ mol-1 (C) C4H10(g) + (13/2) O2(g) -arrow 4CO2(g) + 5H2 O(l) Δc H = - 2658.0 kJ mol-1 (D) C4H10 (g) + (13/2) O2 (g) arrow 4CO2 (g) + 5H2O(l) Δc H = + 2658.0 kJ mol-1

Tardigrade Admin · Accepted Answer

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

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

$2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{2} O (l) Δ_{c} H = - 2658.0 k J m o l^{- 1}$

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (g) Δ_{c} H = - 1329.0 k J m o l^{- 1}$

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) - \to 4 C O_{2} (g) + 5 H_{2} O (l) Δ_{c} H = - 2658.0 k J m o l^{- 1}$

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l) Δ_{c} H = + 2658.0 k J m o l^{- 1}$

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

2C4​H10​(g)+13O2​(g)→8CO2​(g)+10H2​O(l)Δc​H=−2658.0kJmol−1

C4​H10​(g)+213​O2​(g)→4CO2​(g)+5H2​O(g)Δc​H=−1329.0kJmol−1

C4​H10​(g)+213​O2​(g)−→4CO2​(g)+5H2​O(l)Δc​H=−2658.0kJmol−1

C4​H10​(g)+213​O2​(g)→4CO2​(g)+5H2​O(l)Δc​H=+2658.0kJmol−1

C4​H10​(g)+213​O2(g)​⟶4CO2(g)​+5H2​O(1)​; Δc​H=−2658.0kJ/mol The complete combustion of 1 mole of butane is represented by C4​H10(g)​+213​O2(g)​⟶4CO2(g)​+5H2​O(1)​; Δc​H=−2658.0kJ/mol Δc​H should be negative and have a value of 2658kJ/mol.

$2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{2} O (l) Δ_{c} H = - 2658.0 k J m o l^{- 1}$

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (g) Δ_{c} H = - 1329.0 k J m o l^{- 1}$

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) - \to 4 C O_{2} (g) + 5 H_{2} O (l) Δ_{c} H = - 2658.0 k J m o l^{- 1}$

$C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l) Δ_{c} H = + 2658.0 k J m o l^{- 1}$