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Lattice Energy

The standard ...

Question

The standard enthalpy of formation of propene, $C_{3} H_{6}$ is +20.6 kJ mol $^{- 1}$ . The heats of formation of $C O_{2} (g)$ and $H_{2} O (l)$ are -394 kJ mol $^{- 1}$ . and -285.8 kJ mol $^{- 1}$ .

Calculate the heat of combustion of one mole of $C_{3} H_{6}$ .

A

1721.2 kJ mol

^{- 1}

.

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B

-1978.1 kJ mol

^{- 1}

.

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C

2060.0 kJ mol

^{- 1}

.

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D

2221.6 kJ mol

^{- 1}

.

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Solution

The correct option is A 2060.0 kJ mol $^{- 1}$ .
$C_{3} H_{6} + \frac{9}{2} O_{2} ⟶ 3 C O_{2} + 3 H_{2} O$

$3 C + 3 H_{2} ⟶ C_{3} H_{6} Δ H =$ 20.6 kJ mol $^{- 1}$ .

$C + O_{2} ⟶ C O_{2} Δ H = -$ 394 kJ mol $^{- 1}$ .

$H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O Δ H = -$ 285.8 kJ mol $^{- 1}$ .

${(Δ H_{C})}_{C_{3} H_{6}} = [3 Δ H_{C O_{2}} + 3 Δ H_{f (H_{2} O)} - Δ H_{f (C_{3} H_{6})}]$

$= [3 \times (- 394) - 3 (285.8) - 20.6]$

${(Δ H_{C})}_{C_{3} H_{6}} =$ $- 2060$ kJ mol $^{- 1}$ .

Hence, the correct option is $C$ .

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