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Enthalpy of Combustion

The heat evol...

Question

The heat evolved during the combustion of 112 litre of water gas at SPT(mixture of equal volume of $H_{2}$ and CO) is : Given
$H_{2} (g) + 1 / 2 O_{2} (g) = H_{2} O (g); △ H = - 241.8 k J$
$C O (g) + 1 / 2 O_{2} (g) = {C O}_{2} (g); △ H = - 283 k J$

241.8 KJ

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283 KJ

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1312 KJ

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1586 KJ

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Solution

The correct option is B 1312 KJ
$112$ litre gas water ${\begin{matrix} 56 l i t . H_{2} 56 l i t C O \end{matrix}$
By combustion of
$22.4$ lit of $H_{2} \to 241.8 k J$
$56$ lit of $H_{2} \to \frac{241.8 \times 56}{22.4}$
$= 604.5 k J$
By combustion of $22.41 C O \to 2.83 k J$
$∴ 56$ lit $C O = \frac{283 \times 56}{22.4}$
$= 707.5 k J$
Total heat $= 707.5 + 604.5$
$= 1.312 k J$

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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

112

H_{2} (g)

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

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (g); Δ H = - 241.8 k J C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) = 2 C O_{2} (g) + H_{2} O (g); Δ H = - 1300 k J \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

C_{2} H_{2}

H_{2}

C_{3} H_{8}

C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 C O_{2} (g) + 4 H_{2} O (g)

3 C (s) + 4 H_{2} (g) ⟶ C_{3} H_{8} (g) Δ H / k J = - 103.8

C (s) + O_{2} (g) ⟶ C O_{2} (g) Δ H / k J = - 393.5

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

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