InterviewSolution
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InterviewSolution
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On the basis of the following thermochemical data `:` `(Delta_(f)G^(@)H_((aq.))^(+)=0)` `H_(2)O_((l))rarr H_((aq.))^(+)+OH_((aq.))^(-),DeltaH=57.32kJ` `H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)),DeltaH=-286.20kJ` The value of enthalpy of formation of `OH^(-)` ion at `25^(@)C` is `:`A. `-228.88 KJ`B. `+228.88 KJ`C. `-343.52 KJ`D. `-22.88 KJ` |
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Answer» Correct Answer - A `H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)O(l)" " =-286.20 KJ` `DeltaH_(r)=DeltaH_(f)(H_(2)O,l) = DeltaH_(f)(H_(2),g) -(1)/(2)DeltaH_(r) (O_(2),g)` `-286.20 = DeltaH_(f)(H_(2)O(l))` So `DeltaH-=_(f) (H_(2)O,l) =-286.20 KJ//"mole"` `H_(2)O(l) rarr H^(+) (aq) + OH^(-)(aq)" "DeltaH = 57.32 KJ` `DeltaH_(r)= DeltaH_(f)^(@) (H^(+),aq) + DeltaH_(f)^(@)(OH^(-),aq) -DeltaH_(f)^(@)(H_(2)O,l)` `57.32 = 0+ DeltaH_(f)^(@)(OH^(-),aq) -(-286.20)` `DeltaH_(f)^(@)(OH^(-), aq) = 57.32 - 286.20 =-228.88 KJ`. |
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