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Enthalpy of Combustion

The products ...

Question

The products of combustion of an aliphatic thiol (RSH) at $298 K$ are

C O_{2} (g), H_{2} O (g) a n d S O_{2} (g)

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C O_{2} (g), H_{2} O (l) a n d S O_{2} (g)

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C O_{2} (l) . H_{2} O (l) a n d S O_{2} (g)

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C O_{2} (g), H_{2} O (l) a n d S O_{2} (l)

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Solution

The correct option is B $C O_{2} (g), H_{2} O (l) a n d S O_{2} (g)$
Let us assume R to be $C_{2} H_{5}$
$C_{2} H_{5} S H + \frac{9}{2} O_{2} (g) ⟶ 2 C O_{2} (g) + 3 H_{2} O (l) + S O_{2} (g)$
Hence, products are $C O_{2}, H_{2} O a n d S O_{2}$
Since, combustion is taking place at room temperature, $H_{2} O$ will be in liquid state.
Hence, option B is correct

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C H_{4}

C H_{4} (g) + 2 O_{2} (g) ⟶ C O_{2} (g) + 2 H_{2} O (g)

{C H}_{3} O H (l)

{C H}_{3} O H (l) + \frac{3}{2} O_{2} (g) ⟶ C O_{2} (g) + 2 H_{2} O (l); Δ H = - 726 k J

C (s) + O_{2} (g) ⟶ C O_{2} (g); Δ H = - 394 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 286 k J

C H_{3} O H (l)

C H_{3} O H (l) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l); △_{r} H^{⊖} = - 726 k J

C (s) + O_{2} (g) \to C O_{2} (g), Δ H = - 394 k J

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