Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

answer:- **Given:** \(\Delta_{\text{fus}}H^\circ\) (ice) = 6.01 kJ mol\(^{-1}\) at 0 °C, \(\Delta_{\text{vap}}H^\circ\) (H\(_2\)O) = 40.7 kJ mol\(^{-1}\) at 100 °C, Specific heat of water is 4.18 J g\(^{-1}\) K\(^{-1}\) **To find:** The total heat required to carry out the given reaction using 180 g of ice. **Calculation:** \[ \begin{align*} \ce{H2O_{(s)}} &\rightarrow_{\text{[Latent heat]}} \ce{H2O_{(l)}} &&\text{(of fusion 0 °C)} \\ &\rightarrow_{\text{[Heating]}} \ce{H2O_{(l)}} &&\text{(water at 0 °C)} \\ &\rightarrow_{\text{[Latent heat]}} \ce{H2O_{(g)}} &&\text{(of vaporization 100 °C)} \end{align*} \] a) \(\ce{H2O_{(s)}} \rightarrow \ce{H2O_{(l)}}\) - 0 °C \hspace{0.5cm} 0 °C - Heat required = Latent heat for 180 g. 1 mol of H\(_2\)O = 6.01 kJ 1 mol of H\(_2\)O = 18 g \(\therefore\) 180 g of H\(_2\)O = \(\frac{180 \text{ g}}{18 \text{ g mol}^{-1}}\) = 10 moles of H\(_2\)O \(\therefore\) 10 mol of H\(_2\)O requires = 60.1 kJ \(\therefore\) Heat required = 60.1 kJ ...(i) b) \(\ce{H2O_{(l)}} \rightarrow \ce{H2O_{(s)}}\) - 0 °C \hspace{0.5cm} 100 °C - Heat required = Mass \(\times\) Specific heat \(\times\) \(\Delta T\) \(= 180 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 100 \text{ K}\) \(= 75240 \text{ J}\) \(= 75.240 \text{ kJ}\) ...(ii) c) \(\ce{H2O_{(l)}} \rightarrow \ce{H2O_{(g)}}\) - 100 °C \hspace{0.5cm} 100 °C - Heat required = Latent heat of vaporization 1 mol of H\(_2\)O requires = 40.7 kJ \(\therefore\) 1 mol of H\(_2\)O = 18 g \(\therefore\) 180 g of H\(_2\)O = 10 moles of H\(_2\)O \(\therefore\) Heat required by 10 moles of water = 407 kJ ...(iii) From (i), (ii), and (iii), Total heat required to carry out the given reaction using 180 g of ice \(= 60.1 \text{ kJ} + 75.240 \text{ kJ} + 407 \text{ kJ} = +542.34 \text{ kJ}\) The total heat required to melt 180 g of ice at 0 °C, heat it to 100 °C, and then vaporize it at that temperature is +542.34 kJ.