JEE Main On 8 April 2017 Question 28
Question: The enthalpy change on freezing of 1 mol of water at $ 5{}^\circ C $ to ice at $ -5{}^\circ C $ is : (Given $ {\Delta _{fus}}H=6kJ,mo{l^{-1}} $ at $ 0{}^\circ C $ , $ C _{p}(H _2O,\ell )=75.3J,mo{l^{-1}}{K^{-1}}, $ $ C _{p}(H _2O,s)=36.8J,mo{l^{-1}}{K^{-1}}) $ [JEE Online 08-04-2017]
Options:
A) $ 6.00,kJ,mo{l^{-1}} $
B) $ 5.81,kJ,mo{l^{-1}} $
C) $ 5.44,kJ,mo{l^{-1}} $
D) $ -6.56,kJ,mo{l^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
Specific Heat Capacity of Water: $C_{p, \text { water }}$ is the amount of energy needed to raise the temperature of 1 mole of water by $1^{\circ} \mathrm{C}$. Here, it’s $75.3 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$.
The formula to calculate the energy change (Cooling Water from $5^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$) is:
$$ \begin{gathered} \Delta H_1=\text { moles } \times C_{p, \text { water }} \times \Delta T \ \Delta H_1=1 \times 75.3 \times(-5)=-376.5 \mathrm{~J} / \mathrm{mol} \end{gathered} $$
This negative sign indicates that energy is released as the water cools.
- Enthalpy of Fusion ( $\Delta_{\text {fus }} H$ ): This is the energy required to change 1 mole of water at $0^{\circ} \mathrm{C}$ to ice at $0^{\circ} \mathrm{C}$ without changing the temperature. It’s given as $6000 \mathrm{~J} / \mathrm{mol}$.
Since freezing releases energy:
$$ \Delta H_2=-\Delta_{\text {fus }} H=-6000 \mathrm{~J} / \mathrm{mol} $$
Specific Heat Capacity of Ice: $C_{p, \text { ice }}$ is $36.8 \mathrm{~J} / \mathrm{mol} / \mathrm{K}$.
- Temperature Change $(\Delta T)$ : From $0^{\circ} \mathrm{C}$ to $-5^{\circ} \mathrm{C}$ is a decrease of 5 degrees.
The energy change (Cooling Ice from $0^{\circ} \mathrm{C}$ to $-5^{\circ} \mathrm{C}) is:
$$ \begin{gathered} \Delta H_3=\text { moles } \times C_{p, \text { ice }} \times \Delta T \ \Delta H_3=1 \times 36.8 \times(-5)=-184.0 \mathrm{~J} / \mathrm{mol} \end{gathered} $$
Again, the negative sign indicates energy is released. Total Enthalpy Change
$$ \Delta H_{\text {total }}=\Delta H_1+\Delta H_2+\Delta H_3 $$
$$ \Delta H_{\text {total }}=-376.5 \mathrm{~J} / \mathrm{mol}+(-6000 \mathrm{~J} / \mathrm{mol})+(-184.0 \mathrm{~J} / \mathrm{mol})=-6550.5 \mathrm{~J} / \mathrm{mol} $$
Convert this to kilojoules:
$$ \Delta H_{\mathrm{total}}=-6.5505 \mathrm{~kJ} / \mathrm{mol} \approx-6.56 \mathrm{~kJ} / \mathrm{mol} $$
This means that when 1 mole of water is cooled from $5^{\circ} \mathrm{C}$ to ice at $-5^{\circ} \mathrm{C}$, it releases about 6.56 kJ of energy.
BETA
