Given:(I) H_{2(g)} + frac{1}{2}O_{2(g)} to H | vedclass

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(C) The molar enthalpy of vaporisation is the enthalpy change for the process: $H_2O_{(l)} 	o H_2O_{(g)}$.We are given:$(I) \ H_{2(g)} + \frac{1}{2}O

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$44.1$

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(C) The molar enthalpy of vaporisation is the enthalpy change for the process: $H_2O_{(l)} 	o H_2O_{(g)}$.We are given:$(I) \ H_{2(g)} + \frac{1}{2}O_{2(g)} 	o H_2O_{(l)}; \Delta H^o_1 = -285.9 \ kJ \ mol^{-1}$$(II) \ H_{2(g)} + \frac{1}{2}O_{2(g)} 	o H_2O_{(g)}; \Delta H^o_2 = -241.8 \ kJ \ mol^{-1}$To obtain the target equation,subtract equation $(I)$ from equation $(II)$:$(II) - (I): H_2O_{(l)} 	o H_2O_{(g)}$$\Delta H^o_{vap} = \Delta H^o_2 - \Delta H^o_1$$\Delta H^o_{vap} = -241.8 - (-285.9) \ kJ \ mol^{-1}$$\Delta H^o_{vap} = 44.1 \ kJ \ mol^{-1}$

Given:
$(I) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}; \Delta H^o_{298\ K} = -285.9 \ kJ \ mol^{-1}$
$(II) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H^o_{298\ K} = -241.8 \ kJ \ mol^{-1}$
The molar enthalpy of vaporisation of water will be $kJ \ mol^{-1}$.

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Given:$(I) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}; \Delta H^o_{298\ K} = -285.9 \ kJ \ mol^{-1}$$(II) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H^o_{298\ K} = -241.8 \ kJ \ mol^{-1}$The molar enthalpy of vaporisation of water will be $kJ \ mol^{-1}$.