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Enthalpy of Combustion

Which change ...

Question

Which change occurs with the largest increase in entropy at $25^{\circ} C$ ?

B r_{2} (l) \to B r_{2} (g)

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C_{(g r a p h i t e)} \to C_{(d i a m o n d)}

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H_{2} O (s) \to H_{2} O (l)

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H C l (g) + H_{2} O (l) \to H_{3} O^{+} (a q) + C l^{-} (a q)

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Solution

The correct option is A $B r_{2} (l) \to B r_{2} (g)$
Solids have fewest microstates thus the lowest entropy $.$ Liquid have more microstates and thus higher entropy $.$ When a substance is gas it has many more microstates and thus have higest entropy $.$
Hence $,$ option $(A)$ is correct $.$

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H C l (g) + H_{2} O (l) \to H_{3} O^{+} (a q) + C l^{-} (a q)

Δ_{f} H^{o} in kJ

\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g); Δ_{f} H^{o} = - 92.4 kJ H C l (g) + H_{2} O (l) \to H_{3} O^{+} + (a q) + C l^{-} (a q); Δ H^{o} = - 74.8 kJ

Δ_{f} H^{o} of H^{+} (a q) = 0.0 kJ

H_{2} S O_{4} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + H S O_{4}^{-} (a q)

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H_{2} O_{(l)} + H_{2} O_{(l)} ⇌ H_{3} O_{(a q)}^{+} + O H_{(a q)}^{-}

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H C_{2} O_{5} - (a q) + H_{2} O (l) ⇌ H_{3} O + (a q) + C_{3} O_{4}^{2} - (a q)

K_{c l} = 5.9 \times 10^{- 2}

K_{c 2} = 6.4 \times 10^{- 5}

25^{\circ} C

K_{c}

(3)

H_{2} C_{2} O_{4} (a q) + 2 H_{2} O (l) ⇌ 2 H_{3} O + (a q) + C_{2} O_{4}^{2} - (a q)

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