For the combustion of n-octane C8H18 + O2 → CO2 + H2O at 25°C (ignoring resonance in CO2)

Question

For the combustion of n-octane C8H18 + O2 → CO2 + H2O at 25°C (ignoring resonance in CO2) (A) Δ H= Δ E-5.5 × 8.31 × 0.298 in kJ/mol (B) Δ H = Δ E+ 4.5 × 8.31 × 0.298 in kJ/mol (C) Δ H = Δ E - 4.5 × 8.31 × 0.298 in kJ/mol (D) Δ H = Δ E - 4.5 + 8.31 × 0.298 in kJ/mol

Tardigrade Admin · Accepted Answer

C8H18(l)+12.5 O2(g)→8CO2(g)+9H2O(l) Δ H=Δ E+(8-12.5)×8.31×0.298 in kJ/mol

Q. For the combustion of n-octane
$C_{8} H_{1} 8 + O_{2} \to C O_{2} + H_{2} O$ at $2 5^{\circ} C$ (ignoring resonance in $C O_{2}$ )

$Δ H = Δ E - 5.5 \times 8.31 \times 0.298 in k J / m o l$

$Δ H = Δ E + 4.5 \times 8.31 \times 0.298 in k J / m o l$

$Δ H = Δ E - 4.5 \times 8.31 \times 0.298 in k J / m o l$

$Δ H = Δ E - 4.5 + 8.31 \times 0.298 in k J / m o l$

$_{C_{8} H_{18} (l)} + 12.5 O_{2_{(g)}} \to 8 C O_{2_{(g)}} + 9 H_{2} O_{(l)}$
$Δ H = Δ E + (8 - 12.5) \times 8.31 \times 0.298$ in $k J / m o l$

Q. For the combustion of n-octane C8​H1​8+O2​→CO2​+H2​O at 25∘C (ignoring resonance in CO2​)

ΔH=ΔE−5.5×8.31×0.298inkJ/mol

ΔH=ΔE+4.5×8.31×0.298inkJ/mol

ΔH=ΔE−4.5×8.31×0.298inkJ/mol

ΔH=ΔE−4.5+8.31×0.298inkJ/mol