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Question
Derive a relation between ΔH and ΔU for a chemical reaction. Draw neat labelled diagram of calomel electrode. Resistance and conductivity of a cell containing 0.001 M KCI solution at 298K are 1500Ω and 1.46x10-4 S.cm-1 respectively.
Solution
Relation between ΔH and ΔU
The heat of reaction is given by enthalpy change
ΔH=H2 - H1
by defination H= U+PV
H1=U1+P1V1
H2=U2+P2V2
ΔH = (U1+P1V1) - (U2+P2V2)
=(U2-U1)+(P2V2-P1V1)
=ΔU+(P2V2-P1V1)(∵ΔU=U2-U1)
since PV=nRT
for initial state P1V1=n1RT
for final state P2V2=n2RT
P2V2-P1V1=n2RT-n1RT
=(n2-n)RT
=ΔnRT
where Δn = number of moles of gaseous product - number of moles of gaseous reactant.
ΔH=ΔU+ΔnRT |
Diagram : Standard calomel electrode
Data: R 1500Ω
C=0.001M
k= 1.46x10-4Scm-1
To find: b = ?
Solution : `k=b/R`
b=kxR
=1.46x10-4x1500
b=2190x10-4 cm-1
b = 0.219cm-1 |
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