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(a) The measured heats of neutralization...

(a) The measured heats of neutralization of acetic acid, formic acid, hydrocyanic acid, and hydrogen sulphide are 13.20, 13.40, 2.90 and 3.80 KCal per g.equiv. respectively. Arrange these acids in a decreasing order of strength.
(b) Heat of neutralization of formic acid by `NH_(4)OH` is 11.9 KCal per g.equiv. What is the heat of ionization of `NH_(4)OH`?

Text Solution

Verified by Experts

`DeltaH_("(neutralization)")=DeltaH_("(ionization)")+DeltaH_((H^(+)+OH^(-)))`
`therefore DeltaH_("(ionization)")=DeltaH_("(neutralization)")-DeltaH_((H^(+)+OH^(-)))`
`57.32KJ = 13.2 Kcal`
`DeltaH_("(neutralization)")" for all acids have a "-ve" sign."`
`=-13.20-(-13.70)`
`=+0.50"Cal/g.equiv"`
`DeltaH` for ionization of formic acid
`=-13.40+13.70`
`=+0.30" Cal/g.equiv."`
`DeltaH` for ionization of hydrocyanic acid
`=-2.90+13.70`
`=+10.80"kCal/g.equiv"`
`DeltaH` for ionisation of hydrogen sulphide
`=-3.80+13.70`
`=+9.90" KCal/g.equiv"`.
The acid with the lowest positive value of heat of ionization will be the strongest acid. Thus formic acid is the strongest and hydrocyanic acid the weakest acid. The trend in decreasing strength of acids is :
Formic acid > acetic acid ? hydrocyanic acid > hydrogen sulphide.
(b) Thermochemical equations are :
(1) `HCOOH+NH_(4)OHrarr HCOONH_(4)+H_(2)O" "DeltaH_(1)=-11.9KCal`
(2) `HCOOH rarr HCOO^(0)+H^(+)" "DeltaH_(2)=+0.30KCal.`
(3) `NH_(4)OH rarr NH_(4)^(+)+OH^(-)." "DeltaH_(3)=x. KCal`
(4) `H^(+)+OH^(-)rarr H_(2)O_((l))," "DeltaH_(4)=-13.70KCal`
The sum of reaction (2), (3) and (4) should equal reaction (1).
`therefore DeltaH_(1)=DeltaH_(2)+DeltaH_(3)+DeltaH_(4)`
`-11.90=0.30+x-13.70` .
`=+1.50"KCal.g.equiv"^(-1)`
`DeltaH` for ionization of
`NH_(4)OH=+1.50" kCal/g.equiv"^(-1)`
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