The enthalpy changes on freezing of 1 mol of water at $5^{\circ} C$ to ice at $- 5^{\circ} C$ is:
(Given $△_{f u s} H = 6 K J m o l^{- 1}$ at $0^{\circ} C$
$C_{P} (H_{2} O, l) = 75.3 J m o l^{- 1} K^{- 1}$
$C_{P} (H_{2} O, S) = 36.8 J m o l^{- 1} K^{- 1}$ )
a.) $5.81 K J m o l^{- 1}$
b.) $6.56 K J m o l^{- 1}$
c.) $6.00 K J m o l^{- 1}$
d.) $5.44 K J m o l^{- 1}$

Question

The enthalpy changes on freezing of 1 mol of water at $5^{\circ} C$ to ice at $- 5^{\circ} C$ is:
(Given $△_{f u s} H = 6 K J m o l^{- 1}$ at $0^{\circ} C$
$C_{P} (H_{2} O, l) = 75.3 J m o l^{- 1} K^{- 1}$
$C_{P} (H_{2} O, S) = 36.8 J m o l^{- 1} K^{- 1}$ )
a.) $5.81 K J m o l^{- 1}$
b.) $6.56 K J m o l^{- 1}$
c.) $6.00 K J m o l^{- 1}$
d.) $5.44 K J m o l^{- 1}$

Vedantu Content Team · Accepted Answer

Hint:  \tThe enthalpy of water is changing from ${{5}^{\circ }}C$ to $-{{5}^{\circ }}C$. So, enthalpy would be changing three times. First time would be when water is cooled from  to $${{0}^{\circ }}C$$, second time would be fusion of liquid at $${{0}^{\circ }}C$$ at the same temperature and third time would be by changing the temperature of ice from $${{0}^{\circ }}C$$ to $-{{5}^{\circ }}C$.The correct answer is B.\n\nStep by Step answer:Step 1 would be to bring down the temperature that cooling down of water from ${{5}^{\circ }}C$ to $${{0}^{\circ }}C$$ So, enthalpy change would be equal to product of number of moles into ${{C}_{P}}({{H}_{2}}O,l)$ which is then multiplied by change in temperature.Therefore,$$enthalpy\text{ }change~=n\times Cp(H_2O,l)\times \Delta T=1\times 75.3J/mol/K\times {{\left( 5-0 \right)}^{\circ }}C=367.5J=0.3765kJ......\left( 1 \right)$$Here, n represents the number of moles of water which are given as 1 , Cp​(H2​O,l) is the specific heat of liquid water which is given in question and ΔT is temperature change and temperature is brought down from ${{5}^{\circ }}C$to $${{0}^{\circ }}C$$ Also 1kJ=1000JStep 2 is Liquid water at $${{0}^{\circ }}C$$ is fused at the same temperature.So the enthalpy change would be equal to the number of moles multiplied by fusion heat.$$The\text{ }enthalpy\text{ }change~=n\Delta fusH=1\times 6kJ/mol=6kJ......\left( 2 \right)$$ Step 3 would be Ice at $${{0}^{\circ }}C$$  is cooled down to ice at $-{{5}^{\circ }}C$So, the enthalpy change would be equal to the number of moles multiplied by specific heat which in turn is multiplied by change in temperature.$$The\text{ }enthalpy\text{ }change~=nCp(H_2O, s)\Delta T=1\times 36.8J/mol/K\times {{\left( 0-\left( -5 \right) \right)}^{\circ }}C=184J=0.184kJ......\left( 3 \right)$$where, n stands the number of moles of ice which is 1, Cp​($H_2​O$, l) is the specific heat of ice which is given in question and ΔT is temperature change which is equal to 5. Also, 1kJ=1000JAdd (1), (2) and (3)Total enthalpy change would be equal to the sum of all the three changes in enthalpy happening during the process.$$The\text{ }enthalpy\text{ }change\text{ }for\text{ }entire\text{ }process~=0.3765kJ+6kJ+0.184kJ=6.56kJmo{{l}^{-1}}$$ The enthalpy change for the entire process would be equal to 6.56kJ/mol which is option B.\nNote: Enthalpy change is the amount of the heat released or absorbed in the reactions at constant pressure.

The enthalpy changes on freezing of 1 mol of water at 5∘C to ice at −5∘C is:(Given △fusH=6KJmol−1 at 0∘CCP(H2O,l)=75.3Jmol−1K−1 CP(H2O,S)=36.8Jmol−1K−1)a.) 5.81KJmol−1 b.) 6.56KJmol−1c.) 6.00KJmol−1d.) 5.44KJmol−1

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