Caculate the enthalpy change on freezing of 1-0 mol of water at -10-0 o C- fus H-6-03kJ mol -1 at 0 o C C P - H 2 Ol - -75-3J mol -1 K -1 C P - H 2 Os - -36-8J mol -1 K -1

Step 1 : Enthalpy change due to change in temperature #160; 10 ^ 0 C to #160; 0 ^ 0 C Δ H _ 1 = C _ P [ H _ 2 O( l ) #160; ] ×Δ T=75.3×( ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Enthalpy

Caculate the ...

Question

Caculate the enthalpy change on freezing of $1.0$ mol of water at ${- 10.0}^{o} C, Δ_{f u s} H = 6.03 k J {m o l}^{- 1}$ at $0^{o} C$
$C_{P} [H_{2} O (l)] = 75.3 J {m o l}^{- 1} K^{- 1}$
$C_{P} [H_{2} O (s)] = 36.8 J {m o l}^{- 1} K^{- 1}$

Open in App

Solution

Suggest Corrections

Similar questions

1

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}

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. Δ_fusH = 6.03 kJ mol^–1 at 0°C.

C_p[H₂O(l)] = 75.3 J mol^–1 K^–1

C_p[H₂O(s)] = 36.8 J mol^–1 K^–1

1.0 m o l

{10.0}^{\circ} C

- {10.0}^{\circ} C

△_{f u s} H = 6.03 K J

m o l^{- 1}

0^{\circ} 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}

1 m o l

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})

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}

Join BYJU'S Learning Program