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Enthalpy of Combustion

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Question

Find out the heat evolved in combustion if $112$ litres(at STP) od water gas (mixture of equal volue of $H_{2} (g)$ and $C O (g)$
$H_{2} (g) + 1 / 2 O_{2} (g) \to H_{2} O (g)$ $Δ H = - 241.8 k J$
$C O (g) + 1 / 2 O_{2} (g) \to C O_{2} (g)$ $Δ H = - 283 k J$

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Solution

$Δ H_{r x n} = \sum Δ H_{c o m b u s t i o n}^{r e a c t a n t} - \sum Δ H_{c o m b u s t i o n}^{p r o d u c t}$
Volume of $H_{2} = 56$ $l i t r e s =$ Volume of $C O$ (given)
Moles of $H_{2} =$ Moles of $C O = \frac{56}{22.4} = 2.5$ $m o l e s$
For $H_{2}$ , $H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (g)$
$Δ H = - 241.8$ $k J / m o l^{- 1}$
$5 \times - 241.8 = 5 \times Δ H_{c o m b u s t i o n}^{H_{2}}$
$Δ H_{c o m b u s t i o n}^{H_{2}} = - 241.8$ $k J / m o l$
For $2.5$ $m o l e s = - 604.5$ $k J$
Similarly,
$Δ H_{c o m b u s t i o n}^{C O_{2}} = - 283$ $k J / m o l$
For $2.5$ $m o l e s = - 707.5$
Total energy released $= - 707.5 - 604.5 = - 1312$ $k J$

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Similar questions

H_{2}

H_{2} (g) + 1 / 2 O_{2} (g) = H_{2} O (g); Δ H = - 241.8 k J

C O (g) + 1 / 2 O_{2} (g) = C O_{2} (g); Δ H = - 283 k J

H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(g)}

Δ H = 241.8 k J

C O_{(g)} + \frac{1}{2} O_{2 (g)} \to C O_{2 (g)}

Δ H = 283 k J

112

H_{2}

C O

Δ H

k J

C (g) + O_{2} (g) \to C O_{2} (g)

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

Δ H = + 131 k J

C O (g) + \frac{1}{2} O_{2} (g) \to {C O}_{2} (g)

Δ H = - 282 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)

Δ H = - 242 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H = x

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = y

H_{2}

H_{2} (g) + 1 / 2 O_{2} (g) = H_{2} O (g); △ H = - 241.8 k J

C O (g) + 1 / 2 O_{2} (g) = {C O}_{2} (g); △ H = - 283 k J

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