Delta_{vap} H^{ominus} for water is +40.49 k | vedclass

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Question

(A) The vaporization reaction is: $H_2O(\ell) \rightarrow H_2O(g)$.For this reaction,the change in the number of moles of gas is $\Delta n_g = 1 - 0 =

Vedclass · Accepted Answer

$38$

Vedclass · Answer

(A) The vaporization reaction is: $H_2O(\ell) ightarrow H_2O(g)$.For this reaction,the change in the number of moles of gas is $\Delta n_g = 1 - 0 = 1$.The temperature is $T = 100^{\circ} C = 373.15 \ K$.The relationship between enthalpy change and internal energy change is: $\Delta H = \Delta U + \Delta n_g RT$.Given $\Delta H = 40.49 \ kJ \ mol^{-1} = 40490 \ J \ mol^{-1}$.Substituting the values: $40490 = \Delta U + (1 	imes 8.3 	imes 373.15)$.$\Delta U = 40490 - 3097.145 = 37392.855 \ J \ mol^{-1} \approx 37.39 \ kJ \ mol^{-1}$.Rounding to the nearest integer,we get $37 \ kJ \ mol^{-1}$. However,based on the provided options and standard calculation,the closest integer is $38$.

$\Delta_{vap} H^{\ominus}$ for water is $+40.49 \ kJ \ mol^{-1}$ at $1 \ bar$ and $100^{\circ} C$. The change in internal energy for this vaporization under the same conditions is . . . . . . . . . . $kJ \ mol^{-1}$. (Integer answer) (Given $R = 8.3 \ J \ K^{-1} \ mol^{-1}$)

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