The enthalpy of the reaction,H_{2(g)} + frac{ | vedclass

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Question

(A) The reaction $H_{2(g)} + \frac{1}{2}O_{2(g)} 	o H_2O_{(g)}$ has enthalpy change $\Delta H_1$.The reaction $H_{2(g)} + \frac{1}{2}O_{2(g)} 	o H_2

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$\Delta H_1 > \Delta H_2$

Vedclass · Answer

(A) The reaction $H_{2(g)} + \frac{1}{2}O_{2(g)} 	o H_2O_{(g)}$ has enthalpy change $\Delta H_1$.The reaction $H_{2(g)} + \frac{1}{2}O_{2(g)} 	o H_2O_{(l)}$ has enthalpy change $\Delta H_2$.The difference between the two reactions is the phase change of water: $H_2O_{(g)} 	o H_2O_{(l)}$.Since condensation is an exothermic process,the enthalpy change for this phase change is negative $(\Delta H Therefore,$\Delta H_2 = \Delta H_1 + \Delta H_{condensation}$.Since $\Delta H_{condensation}  \Delta H_2$.

The enthalpy of the reaction,$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}$ is $\Delta H_1$ and that of $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$ is $\Delta H_2$,then:

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