The enthalpy of formation of H2O(1) 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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- 334 kJ / mol

- 234 kJ / mol

- 346 kJ / mol

- 178 kJ / mol

answer is B.

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Detailed Solution

H2(g)+12O2(g)→H2O(l); ΔH1=−290kJH++OH−→H2O; ΔH2=−56kJ12H2→H+(aq); ΔH=0∴ 12H2(g)+12O2(g)→OH−(aq)ΔrHΔrH=−290+56=−234kJ/mol