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Equipartition Theorem

At 00C,Δ H0...

Question

At $0^{0} C, Δ H_{f u s}^{0}$ = $6 k J / m o l$ , change of entropy for freezing of one mole of ice will be :

\infty

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0

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21.98 J / m o l - K

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13.6 J / m o l - K

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Solution

The correct option is C $21.98 J / m o l - K$
$Δ H_{f} = 6 k J / m o l = 6000 J / m o l$ , $T_{f} = 273 K$

$Δ S = \frac{Δ H_{f}^{0}}{T_{f}} = \frac{6000}{273} = 21.98 J / m o l - K$

Suggest Corrections

Similar questions

1

5^{o} C

- 5^{o} C

Δ_{f u s} H = 6 k J {m o l}^{- 1}

0^{o} C

C_{P} (H_{2} O, l) = 75.3 J {m o l}^{- 1}

K^{- 1}

C_{P} (H_{2} O, s) = 36.8 J {m o l}^{- 1}

K^{- 1}

5^{o} C

- 5^{o} C

[{G i v e n : △}_{f u s} H = 6 k J {m o l}^{- 1} a t 0^{o} C, C_{p} (H_{2} O, l) = 75.3 J {m o l}^{- 1} K^{- 1}, C_{p} (H_{2} O, s)

= 36.8 J {m o l}^{- 1} K^{- 1}]

H_{2} O (s) ⇌ H_{2} O (l)

6.01 k J m o l^{- 1} .

5^{0} C

- 5^{0} C

(Given $Δ_{f u s} H = 6 k J m o l^{- 1} a t 0^{0} C$ ,

$C_{p} (H_{2} O, 1) = 75.3 J m o l^{- 1} K^{- 1}$

$C_{p} (H_{2} O, s) = 36.8 J m o l^{- 1} K^{- 1})$ )

0^{\circ} C

H_{2} O_{(s)} ⇌ H_{2} O_{(l)} i s 6.0 k J m o l^{- 1}

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