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Q.

The enthalpy of formation of is H2O(l) is -290 kJ/mol and enthalpy of neutralization of strong acid with strong alkali is -56 kJ/equiv. what is the enthalpy of formation of OH-(aq)? Given Δform HH+,aq=0

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a

-334 kJ/mol

b

-234 kJ/mol

c

-346 kJ/mol

d

-178 kJ/mol

answer is B.

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Detailed Solution

H2(g)+12O2(g)→H2O(l);ΔH1=-290kJ H++OH-→H2O;ΔH2=-56 kJ 12H2→H+(aq);ΔH=0 ∴12H2(g)+12O2(g)→OH-(aq)ΔrH ΔrH=-290+56=-234 kJ/mol

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