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The enthalpy of the reaction, H2g+12O2g→H2Og, ∆H1 and that of H2g+12O2→H2Ol, ∆H2. Then the correct between the magnitude of ∆H1 and ∆H2 is

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a

b

c

d

answer is A.

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Detailed Solution

H2g+12O2g→IH2Ol→IIH2Og; Step II is endothermic because it is vapourization; Thus ∆H1∆fHH2Og<∆H2∆fHH2Ol;

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