Which of the following reactions is associated with the most negative change in entropy?

2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)

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C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g)

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C (s, g r a p h i t e) + \to C O_{2} (g)

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3 C_{2} H_{2} (g) \to C_{6} H_{6} (l)

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Solution

The correct option is B $3 C_{2} H_{2} (g) \to C_{6} H_{6} (l)$
$2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)$
3 moles of gas becomes 2 moles of gas
$C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g)$
2 moles of gas becomes 1 mole of gas
$C (s, g r a p h i t e) + O_{2} (g) \to C O_{2} (g)$
1 mole of gas becomes 1 mole of gas
$3 C_{2} H_{2} (g) \to C_{6} H_{6} (l)$
3 moles of gas becomes 1 mole of liquid
In (d) 3 moles of gas becomes 1 mole of liquid. If greater number of molecules transforms into a small number of molecules then entropy decreases. Also in (d) gas molecules changes to liquid, hence entropy decreases.
Hence (d) has the most negative change in entropy

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Similar questions

2 {S O}_{2} (g) + O_{2} (g) \to 2 {S O}_{3} (g)

C H_{4} (g) + 2 S_{2} (g) ⇌ C S_{2} (g) + 2 H_{2} S (g)

C O_{2} (g) + C (s) ⇌ 2 C O (g)

4 N H_{3} (g) + 5 O_{2} (g) ⇌ 4 N O (g) + 6 H_{2} O (g)

C_{2} H_{4} (g) + H_{2} (g) ⇌ C_{2} H_{6} (g)

C O C l_{2} (g) ⇌ C O (g) + C l_{2} (g)

C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g), Δ H = + Q c a l

S O_{3}

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) ⟶ 2 {C O}_{2} (g) + H_{2} O (l); △ G^{0} = 1234 k J

C (s) + O_{2} (g) ⟶ {C O}_{2} (g); △ G^{0} = - 394 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); △ G^{0} = - 237 k J

H_{2} (g) + 2 C (s) \to C_{2} H_{2} (g)

6 C (s) + 3 H_{2} (g) ⟶ C_{6} H_{6} (l), Δ H = 49.0 k J

H_{2} (g) + 1 / 2 O_{2} (g) ⟶ H_{2} O (l), Δ H = - 285.8 k J

C (s) + O_{2} (g) ⟶ C O_{2} (g), Δ H = - 389.3 k J

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