For which of the following, there is increase in entropy?

N a (s) + H_{2} I (I) \to N a O H (a q) + H_{2} (g)

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I_{2} (g) \to I_{2} (s)

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H_{2} {S O}_{4} (a q) + {B a (O H)}_{2} (a q) \to B a {S O}_{4} (s) + H_{2} O (I)

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H_{2} (g) + 1 / 2 O_{2} (g) \to H_{2} O (I)

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None of the above

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Solution

The correct option is A $N a (s) + H_{2} I (I) \to N a O H (a q) + H_{2} (g)$
A. $N a (s) + H_{2} I (l) ⟶ N a O H (a q) + H_{2} (g)$
In this equation, the reactants are in solid state and liquid state whereas product is converted to gas.
Thus, since gases have least Intermolecular forces, thus the particles have more randomness.
Thus, here from solid state, product formed is a gas. So, there is increase in entropy.

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Similar questions

H_{(a q)}^{+} + O H_{(a q)}^{-} = H_{2} O_{(I)}, △ H = - X_{1} k J m o l^{- 1}

H_{2 (g)} + \frac{1}{2} O_{2 (g)} = H_{2} O_{(I)}, △ H = - X_{2} k J m o l^{- 1}

C O_{2 (g)} + H_{2 (g)} = C O_{(g)} + H_{2} O_{(I)} - X_{3} k J m o l^{- 1}

C_{2} H_{2 (g)} + \frac{5}{2} O_{2 (g)} = 2 C O_{2 (g)} + H_{2} O_{(I)} + X_{4} k J m o l^{- 1}

H_{2} O_{(l)}

H_{2} (g) + I_{2} (g) \to 2 H I (g) = k [H_{2}] [I_{2}]

For which of the following equation will $Δ H$ be equal to $Δ U$ .

H_{2} (g) + I_{2} (s) \to 2 H I (g); Δ H = 51.9 K J / m o l e

H_{2} (g) + I_{2} (g) \to 2 H I (g); Δ H = - 9.2 K J / m o l e

The heat of reaction of $I_{2} (s) \to I_{2} (g)$ is:

A l (s) + H_{2} {S O}_{4} (a q) ⟶ {A l}_{2} {({S O}_{4})}_{3} (a q) + H_{2} (g)

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