When 2 mole of $C_{2} H_{6}$ are completely burnt, 3129 KJ of heat is liberated. Calculate the heat of formation of $C_{2} H_{6} . △ H_{f}$ for $C O_{2} & H_{2} O$ are -395 KJ and -286 KJ respectively.

Q. When

2

mole of

C_{2} H_{6}

are completely burnt,

3129 k J

of heat is liberated. Calculate the heat of formation of

C_{2} H_{6}

Δ H_{f}

for

{C O}_{2}

and

H_{2} O

are

- 395 k J

and

- 286 k J

respectively.

Q. The

Δ_{f} H^{o}

for the

C O_{2} (g), C O (g)

and

H_{2} O (l)

are -393.5, -110.5 and -241.8

k J m o l^{- 1}

respectively. The standard enthalpy change (in

k J m o l^{- 1}

) for the given reaction is :

C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (l)

Join BYJU'S Learning Program

When 2 moles of C2H6(g) are completely burnt 3120 kJ of heat is liberated. The enthalpy offormation, of C2H6(g) in kJ / mol is X. Then -X is Given ΔH for CO2(g) & H2O(l) are −395 & −285kJ respectively.

C2H6(g)+72O2(g)→2CO2(g)+3H2O(l) Here heat of combustion for two mole of C2H6 =3120 kj , So for one mole of C2H6 =1560 kj Now we know that ΔH0reaction=ΣΔH0fProduct−ΣΔH0fReactant Now using this formula -1560 = (−2×395−3×285)−X So X = -85 KJ / mol -X = 85

When 2 moles of $C_{2} H_{6} (g)$ are completely burnt 3120 kJ of heat is liberated. The enthalpy of
formation, of $C_{2} H_{6} (g)$ in kJ / mol is X. Then -X is
Given $Δ H$ for $C O_{2} (g)$ & $H_{2} O (l)$ are $- 395$ & $- 285 k J$ respectively.