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Entropy

In which of t...

Question

In which of the following reactions, standard reaction entropy change $(△ S^{0})$ is positive and standard Gibb's energy change $(△ G^{0})$ decreases sharply with increasing temperature?

C_{(g r a p h i t e)} + \frac{1}{2} O_{2 (g)} \to {C O}_{(g)}

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

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

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\frac{1}{2} C_{(g r a p h i t e)} + \frac{1}{2} O_{2 (g)} \to \frac{1}{2} {C O}_{2 (g)}

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Solution

The correct option is A $C_{(g r a p h i t e)} + \frac{1}{2} O_{2 (g)} \to {C O}_{(g)}$
The change in gaseous moles $Δ n_{g}$ must be highest which indicates the more gaseous product is formed which increases randomness and thus entropy increases.
$(A)$ $Δ n_{g} = 1 - \frac{1}{2} = \frac{1}{2}$
$(B)$ $Δ n_{g} = 1 - \frac{3}{2} = \frac{- 1}{2}$
$(C)$ $Δ n_{g} = 0 - \frac{1}{2} = \frac{- 1}{2}$
$(D)$ $Δ n_{g} = \frac{1}{2} - \frac{1}{2} = 0$

Hence A is correct.

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Similar questions

Δ U^{o}

300 K

4 C O (g) + 8 H_{2} (g) ⟶ 3 {C H}_{4} (g) + {C O}_{2} (g) + 2 H_{2} O (l)

C_{(g r a p h i t e)} + 1 / 2 O_{2} (g) ⟶ C O (g); Δ {H_{1}}^{o} = - 110.5 k J

C O (g) + 1 / 2 O_{2} (g) ⟶ {C O}_{2} (g); Δ {H_{2}}^{o} = - 282.9 k J

H_{2} (g) + 1 / 2 O_{2} (g) ⟶ H_{2} O (l); Δ {H_{3}}^{o} = - 285.8 k J

C_{(g r a p h i t e)} + 2 H_{2} (g) ⟶ {C H}_{4} (g); Δ {H_{4}}^{o} = - 74.8 k J

Determine $Δ U^{\circ}$ at 300 K for the following reaction using the listed enthalpies of reaction.

$4 C O (g) + 8 H_{2} (g) \to 3 C H_{4} (g) + C O_{2} (g) + 2 H_{2} O (l)$

$C (g r a p h i t e) + \frac{1}{2} O_{2} (g) \to C O (g); Δ {H_{1}}^{\circ} = - 110.5 k J$

$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); Δ {H_{2}}^{\circ} = - 282.9 k J$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H_{3}^{0} = - 285.8 k J$

$C (g r a p h i t e) + 2 H_{2} (g) \to C H_{4} (g); Δ H_{4}^{0} = - 74.8 k J$

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)

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

- 0.094 k J m o l^{- 1} K^{- 1} .

C O_{2}

C O

- 394.4 a n d - 137.2 k J m o l^{- 1}

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

300 K

- 0.094 k J {m o l}^{- 1} K^{- 1}

{C O}_{2}

C O

- 394.4

- 137 k J {m o l}^{- 1}

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