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Answer :
Final answer:
The ∆s for the phase change of 1.69 moles of water from solid to liquid at 0°C is 10.1449 kJ.
Explanation:
To determine the ∆s for the phase change of 1.69 moles of water from solid to liquid at 0°C, we can use the formula ∆H = n × ∆H. Here, ∆H is the enthalpy change, which is given as 6.01 kJ/mol, and n is the number of moles of water, which is 1.69 moles. Plugging these values into the formula, we get: ∆H = (1.69 moles) × (6.01 kJ/mol) = 10.1449 kJ. Therefore, the ∆s for the phase change of 1.69 moles of water from solid to liquid at 0°C is 10.1449 kJ.
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Final answer:
The ∆S (change in entropy) for the phase change of 1.69 moles of water from solid to liquid at 0°C can be calculated using the equation ∆S = ∆H/T, where ∆H is the enthalpy change and T is the temperature in Kelvin. In this case, the ∆S is 0.022 J/(mol K).
Explanation:
The enthalpy change (∆H) for the phase change of 1.69 moles of water from solid to liquid at 0°C is given as 6.01 kJ/mol. To determine the ∆S (change in entropy) for this process, we need to know the melting temperature of ice and the ∆H fus of water. The melting temperature of ice is 0°C, which is also the temperature at which the phase change occurs. The ∆H fus for water is 6.01 kJ/mol. Therefore, the ∆S for this phase change can be calculated using the equation ∆S = ∆H/T, where T is the temperature in Kelvin.
∆S = (6.01 kJ/mol) / (0 + 273 K) = 0.022 J/(mol K)
