Question:
What is the enthalpy change \((in~J~mol^{-1})\) for the conversion of 1 mole of \(H_{2}O (l)\) at \(10^{\circ}C\) to 1 mole of \(H_{2}O (s)\) at \(-10^{\circ}C\)? \((At~0^{\circ}C~H_{2}O(s)+x~kj~mol^{-1}\rightarrow H_{2}O(l); C_{p}(H_{2}O(l))=yJ~mol^{-1}K^{-1}; C_{p}(H_{2}O(s))=zJ~mol^{-1}K^{-1})\)
What is the enthalpy change \((in~J~mol^{-1})\) for the conversion of 1 mole of \(H_{2}O (l)\) at \(10^{\circ}C\) to 1 mole of \(H_{2}O (s)\) at \(-10^{\circ}C\)? \((At~0^{\circ}C~H_{2}O(s)+x~kj~mol^{-1}\rightarrow H_{2}O(l); C_{p}(H_{2}O(l))=yJ~mol^{-1}K^{-1}; C_{p}(H_{2}O(s))=zJ~mol^{-1}K^{-1})\)
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Always ensure energy units are consistent (convert kJ to J) and pay close attention to the sign of \(\Delta T\) and the phase change enthalpy when dealing with endothermic or exothermic processes.
Updated On: Jun 8, 2026
- \(-(1000x+10y+10z)\)
- \(-(x + y + z)\)
- \(-(1000x+y-z)\)
- \(-(1000x-y+z)\)
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The Correct Option is A
Solution and Explanation
Concept:
According to Hess's Law, the total enthalpy change for a process is the sum of the enthalpy changes of its individual constituent steps. We divide this transformation into three distinct stages to calculate the overall energy change.
Step 1: Cooling liquid water from \(10^{\circ}C\) to \(0^{\circ}C\).
The enthalpy change for cooling is \(\Delta H_1 = n \cdot C_p(l) \cdot \Delta T\). Substituting the given values: \(\Delta H_1 = 1 \cdot y \cdot (0 - 10) = -10y\) Joules.
Step 2: Phase change from liquid to solid at \(0^{\circ}C\).
The problem defines \(H_2O(s) + x \text{ kJ} \rightarrow H_2O(l)\). Therefore, for the reverse process (liquid to solid), the enthalpy change is \(\Delta H_2 = -x \text{ kJ/mol} = -1000x\) Joules per mole.
Step 3: Cooling ice from \(0^{\circ}C\) to \(-10^{\circ}C\).
The enthalpy change for cooling the solid phase is \(\Delta H_3 = n \cdot C_p(s) \cdot \Delta T\). Substituting the values: \(\Delta H_3 = 1 \cdot z \cdot (-10 - 0) = -10z\) Joules.
Step 4: Calculate total enthalpy change.
The total change \(\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -10y - 1000x - 10z = -(1000x + 10y + 10z)\) J mol\(^{-1}\).
Step 1: Cooling liquid water from \(10^{\circ}C\) to \(0^{\circ}C\).
The enthalpy change for cooling is \(\Delta H_1 = n \cdot C_p(l) \cdot \Delta T\). Substituting the given values: \(\Delta H_1 = 1 \cdot y \cdot (0 - 10) = -10y\) Joules.
Step 2: Phase change from liquid to solid at \(0^{\circ}C\).
The problem defines \(H_2O(s) + x \text{ kJ} \rightarrow H_2O(l)\). Therefore, for the reverse process (liquid to solid), the enthalpy change is \(\Delta H_2 = -x \text{ kJ/mol} = -1000x\) Joules per mole.
Step 3: Cooling ice from \(0^{\circ}C\) to \(-10^{\circ}C\).
The enthalpy change for cooling the solid phase is \(\Delta H_3 = n \cdot C_p(s) \cdot \Delta T\). Substituting the values: \(\Delta H_3 = 1 \cdot z \cdot (-10 - 0) = -10z\) Joules.
Step 4: Calculate total enthalpy change.
The total change \(\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -10y - 1000x - 10z = -(1000x + 10y + 10z)\) J mol\(^{-1}\).
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