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Enthalpy of Solution

Compound Ht ...

Question

Compound $△ H_{t} (k c a l / m o l)$
$H_{2} O (g)$ $- 57.8$
$C_{2} H_{4} (g)$ $12.5$
$C O_{2} (g)$ $- 94.1$
What is the heat of combustion of one mole of $C_{2} H_{4}$ ?

A

+ 316.3 k c a l

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B

- 12.5 k c a l

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C

- 291.3 k c a l

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D

- 374.1 k c a l

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E

- 57.8 k c a l

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Solution

The correct option is C $- 374.1 k c a l$
The heat of combustion for $C_{2} H_{4}$ reaction is :
$C_{2} H_{6} + \frac{7}{2} O_{2} \to 2 C O_{2} + 3 H_{2} O$
$Δ_{r e a c t i o n} = Δ H_{f}$ (products)- $Δ H_{f}$ (reactants)
$= Δ H_{f} (2 C O_{2} + 3 H_{2} O) - Δ H_{f} (C_{2} H_{6} + \frac{7}{2} O_{2})$
$= 2 Δ H_{f} C O_{2} + 3 Δ H_{f} H_{2} O - Δ H_{f} C_{2} H_{6} - \frac{7}{2} Δ H_{f} O_{2} = 2 (- 94.1) + 3 (- 57.8) - (12.5) - 0 = - 374.1 k c a l / m o l$

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Similar questions

Q. Calculate the heat of combustion of

1

C_{2} H_{4} (g)

{C O}_{2} (g)

H_{2} O (g)

398 K

1

atmosphere, given that the heats of formation of

{C O}_{2} (g), H_{2} O (g)

C_{2} H_{4} (g)

- 94.1, - 57.8

+ 12.5

k c a l

{m o l}^{- 1}

Q. Using the table given below, determine the heat of combustion of one mole of

C_{2} H_{4}

at 298 K and 1 atm?

Compound	$Δ$ of formation (kcal/mol)
$H_{2} O (g)$	-57.8
$C_{2} H_{6} (g)$	-20/2
$C_{2} H_{4} (g)$	12.5
$C_{2} H_{2} (g)$	54.2
$C O (g)$	-26.4
${C O}_{2} (g)$	-94.1

Q. The standard heats of formation at

298 K

C {C l}_{4} (g), H_{2} O (g), {C O}_{2} (g)

H C l (g)

- 25.5, - 57.8, - 94.1

- 22.1

k c a l

{m o l}^{- 1}

respectively. Calculate

Δ H_{298}^{o}

for the reaction.

C {C l}_{4} (g) + H_{2} O (g) ⟶ {C O}_{2} (g) + 4 H C l (g)

Q. The heat of formation of

C O_{2} (g)

is -394 kJ/mole and that of

H_{2} O (l)

is -286 kJ/mole. The heat of combustion of

C_{2} H_{4}

is -1412 kJ/mole. What is the heat of formation of

C_{2} H_{4}

C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l)

Q. The standard enthalpy of formation of CH4(g), CO2(g) and H2O(g) are -76.2, -394.8 and -241.6 kJ/mol respectively. Calculate the heat of combustion of methane.

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