The standard heat of combustion of propane is -2220.1 kJ mol-1. The standard heat of vaporisation of liquid water is 44.0 kJ mol-1. What is Δ H° of C3H8(g) + 5O2(g)→ 3CO2(g) + 4H2O(g)

Question

The standard heat of combustion of propane is -2220.1 kJ mol-1. The standard heat of vaporisation of liquid water is 44.0 kJ mol-1. What is Δ H° of C3H8(g) + 5O2(g)→ 3CO2(g) + 4H2O(g) (A) -2220.1 kJ (B) -2044.1 kJ (C) -2396.1 kJ (D) -2176.1 kJ

Tardigrade Admin · Accepted Answer

Δ H°( textvap) of H2O =44.0 kJ mol -1 ∴ Δ H° text(vap) of 4 mol H2O=176.0 kJ ∴ Δ H° textRxn=Δ H° textcomb(C3H8)+Δ H° textvap(4H2O) =-2220.1+176.0 =-2044.1 kJ

Q. The standard heat of combustion of propane is $- 2220.1 k J m o l^{- 1} .$ The standard heat of vaporisation of liquid water is $44.0 k J m o l^{- 1}$ . What is $Δ H^{\circ}$ of
$C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 C O_{2} (g) + 4 H_{2} O (g)$

$- 2220.1 k J$

$- 2044.1 k J$

$- 2396.1 k J$

$- 2176.1 k J$

$Δ H_{(vap)}^{\circ} o f H_{2} O = 44.0 k J$ mol $^{- 1}$
$∴ Δ H_{(vap)}^{\circ}$ of $4 m o l$ $H_{2} O = 176.0 k J$
$∴ Δ H_{Rxn}^{\circ} = Δ H_{comb}^{\circ} (C_{3} H_{8}) + Δ H_{vap}^{\circ} (4 H_{2} O)$
$= - 2220.1 + 176.0$
$= - 2044.1 k J$

Q. The standard heat of combustion of propane is −2220.1kJmol−1. The standard heat of vaporisation of liquid water is 44.0kJmol−1. What is ΔH∘ of C3​H8​(g)+5O2​(g)→3CO2​(g)+4H2​O(g)

−2220.1kJ

−2044.1kJ

−2396.1kJ

−2176.1kJ