From the following data,the enthalpy of vapor | vedclass

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(A) The enthalpy of vaporization is the energy required to convert liquid water to gaseous water: $H_2O(l) \rightarrow H_2O(g)$.We can obtain this rea

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$+43.93$

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(A) The enthalpy of vaporization is the energy required to convert liquid water to gaseous water: $H_2O(l) \rightarrow H_2O(g)$.We can obtain this reaction by subtracting the first equation from the second equation:$(H_2(g) + 1/2 O_2(g)$ $\rightarrow H_2O(g)) - (H_2(g) + 1/2 O_2(g)$ $\rightarrow H_2O(l))$$\Delta H_{vap} = \Delta H_2 - \Delta H_1$$\Delta H_{vap} = (-241.84 \, KJ \, mol^{-1}) - (-285.77 \, KJ \, mol^{-1})$$\Delta H_{vap} = -241.84 + 285.77 = +43.93 \, KJ \, mol^{-1}$.

From the following data,the enthalpy of vaporization of liquid water in $KJ \, mol^{-1}$ will be:
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l); \Delta H = -285.77 \, KJ \, mol^{-1}$
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(g); \Delta H = -241.84 \, KJ \, mol^{-1}$

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From the following data,the enthalpy of vaporization of liquid water in $KJ \, mol^{-1}$ will be:$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l); \Delta H = -285.77 \, KJ \, mol^{-1}$$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(g); \Delta H = -241.84 \, KJ \, mol^{-1}$