At 25° C and 1 atm pressure, the enthalpies of combustion are as given below: Substance H2 C(graphite) C2H6(g) (ΔcH ominus/Kj mol-1) -286.0 -394.0 T-1560.0 The enthalpy of formation of ethane is

Question

At 25° C and 1 atm pressure, the enthalpies of combustion are as given below: Substance H2 C(graphite) C2H6(g) (ΔcH ominus/Kj mol-1) -286.0 -394.0 T-1560.0 The enthalpy of formation of ethane is (A) +54.0 kJ mol -1 (B) -68.0 kJ mol -1 (C) -86.0 kJ mol -1 (D) +97.0 kJ mol -1

Tardigrade Admin · Accepted Answer

C 2 H 6( g )+(7/2) O 2( g ) arrow 2 CO 2( g )+3 H 2 O (l ) Δ C H ( C 2 H 6)=2 Δ f H CO 2( g )+3 Δ f H ( H 2 O , ℓ) -Δ f H ( C 2 H 6, g ) -1560=2(-394)+3(-286)-Δ f H ( C 2 H 6, g ) Δ f H ( C 2 H 6, g )=-86 kJ / mole

Q. At $2 5^{\circ} C$ and $1 a t m$ pressure, the enthalpies of combustion are as given below:

Substance $H_{2}$ C(graphite) $C_{2} H_{6} (g)$

$\frac{Δ _{c} H ^{⊖}}{K j m o l ^{- 1}}$ $- 286.0$ $- 394.0$ T-1560.0

The enthalpy of formation of ethane is

$+ 54.0 k J m o l^{- 1}$

$- 68.0 k J m o l^{- 1}$

$- 86.0 k J m o l^{- 1}$

$+ 97.0 k J m o l^{- 1}$

$C_{2} H_{6} (g) + \frac{7}{2} O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)$
$Δ_{C} H (C_{2} H_{6}) = 2 Δ_{f} H C O_{2} (g) + 3 Δ_{f} H (H_{2} O, ℓ)$
$- Δ_{f} H (C_{2} H_{6}, g)$
$- 1560 = 2 (- 394) + 3 (- 286) - Δ_{f} H (C_{2} H_{6}, g)$
$Δ_{f} H (C_{2} H_{6}, g) = - 86 k J / m o l e$

Q. At 25∘C and 1atm pressure, the enthalpies of combustion are as given below: Substance H2​ C(graphite) C2​H6​(g) Kjmol−1Δc​H⊖​ −286.0 −394.0 T-1560.0 The enthalpy of formation of ethane is

+54.0kJmol−1

−68.0kJmol−1

−86.0kJmol−1

+97.0kJmol−1