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Equipartition Theorem

At 0oC, ice...

Question

At $0^{o} C$ , ice and water are in equilibrium and enthalpy change for the process $H_{2} O (s) ⟶ H_{2} O (l)$ is $6 k J {m o l}^{- 1}$ . Calculate the entropy change for the conversion of ice into liquid water.

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Solution

We know that for the process of fusion,
$H_{2} O (s) ⟶ H_{2} O (l)$
$Δ S_{f u s i o n} = \frac{Δ H_{f u s i o n}}{T_{f}} = \frac{6 \times 1000}{273} = 21.98 J K^{- 1} {m o l}^{- 1}$

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Similar questions

0^{\circ} C

, ice and water are in equilibrium and enthalpy change for the process

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

. The entropy change for the conversion of ice into liquid water is

Q. One mole of ice is converted into water at

273

K. The entropies of

H_{2} O (s)

H_{2} O (l)

38.20

60.01

J . m o l^{- 1}

K^{- 1}

respectively. The enthalpy change for the conversion is:

18 g m

of ice is converted into water at

0^{o} C

1

atm. The entropies of

H_{2} O (s)

H_{2} O (I)

38.2

60 J / m o l K

respectively. The enthalpy changes for this conversion is:

Q. One mole of ice is converted into water at

273 K

. The entropies of

H_{2} O (s)

H_{2} O (l)

38.20

J m o l^{- 1}

60.01

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

respectively. The enthalpy of change for the conversion is:

Q. Calculate the enthalpy change of fusion of

1

mol of water at

5^{o} C

- 5^{o} C

Δ_{f u s} H = 6.03 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}

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