Given the following standard heats of reactions:
(a) heat of formation of water $= - 68.3 k c a l$ , (b) heat of combustion of $C_{2} H_{2} = - 310.6 k c a l$ and (c) heat of combustion of ethylene $= - 337.2 k c a l$ . Calculate the heat of the reaction for the hydrogenation of acetylene at constant volume and at $25^{o} C$ .

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Solution

The required equation is
$C_{2} H_{2} (g) + H_{2} (g) \to C_{2} H_{4} (g); Δ H_{f} = ?$
Given
(a) $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); (Δ H = - 68.3 k c a l) . . . . (i)$
(b) $C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 {C O}_{2} (g) + H_{2} O (l); (Δ H = - 310.6 k c a l) . . . . (i i)$
(c) $C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 {C O}_{2} (g) + 2 H_{2} O (l) (Δ H = - 337.2 k c a l) . . . . (i i i)$
The required equation can be achieved by adding eqs. $(i)$ and $(i i)$ and subtracting $(i i i)$
$C_{2} H_{2} (g) + H_{2} (g) + 3 O_{2} (g) - C_{2} H_{4} (g) - 3 O_{2} (g) \to 2 {C O}_{2} + 2 H_{2} O (l) - 2 {C O}_{2} (g) - 2 H_{2} O (l)$
or $C_{2} H_{2} (g) + H_{2} (g) \to C_{2} H_{4} (g)$
$Δ H = - 68.3 - 310.6 - (- 337.2) = - 378 + 337.2 = - 41.7 k c a l$
We know that
$Δ H = Δ U + Δ n R T$
or $Δ U = Δ H - Δ n R T$
or $Δ n = (1 - 2) = - 1, R = 2 \times 10^{- 3} k c a l {m o l}^{- 1} K^{- 1}$
and $T = (25 + 273) = 298 K$
Substituting the values in above equation
$Δ U = - 41.7 - (- 1) (2 \times 10^{- 3}) (298)$
$= - 41.104 k c a l$

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Similar questions

= - 68.3 k c a l

= - 310.6 k c a l

= - 337.2 k c a l

25^{\circ} C

- 68.3 kcal

- 310.6 kcal

- 337.2 kcal

(27^{\circ} C)

R = 2 \times 10^{- 3} kcal K^{- 1} {mol}^{- 1}

- 68.3 k c a l

- 310.6 k c a l

- 337.2 k c a l

(27^{\circ} C)

18^{o} C

- 335.8 k c a l

18^{o} C

- 394 k J, - 286 k J

- 1301 k J

C_{2} H_{2}

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