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calculate the enthalpy change on freezing of 1 0 m...

Question:
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

Answer:

Total enthalpy change involved in the transformation is the sum of the following changes:

(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.

(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

Total ΔH = C_p [H₂OCI] ΔT + ΔH_freezing + C_p[H₂O_(s)] ΔH

= (75.3 J mol^-1 K^-1) (0 - 10)K + (-6.03 × 10³ J mol^-1) + (36.8 J mol^-1 K^-1) (-10 - 0)K

= -753 J mol^-1- 6030 J mol^-1- 368 J mol^-1

= -7151 J mol^-1

= -7.151 kJ mol^-1

Hence, the enthalpy change involved in the transformation is -7.151 kJ mol^-1.

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Question:
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

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Question: Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. ΔfusH = 6.03 kJ mol-1 at 0°C. Cp[H2O(l)] = 75.3 J mol-1 K-1 Cp[H2O(s)] = 36.8 J mol-1 K-1

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Question:
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