#### 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=H_{2} - H_{1}

by defination H= U+PV

H_{1}=U_{1}+P_{1}V_{1}

H_{2}=U_{2}+P_{2}V_{2}

ΔH = (U_{1}+P1V_{1}) - (U_{2}+P_{2}V_{2})

=(U_{2}-U_{1})+(P_{2}V_{2}-P_{1}V_{1})

=ΔU+(P_{2}V_{2}-P_{1}V_{1})(∵ΔU=U_{2}-U_{1})

since PV=nRT

for initial state P_{1}V_{1}=n_{1}RT

for final state P_{2}V_{2}=n_{2}RT

P_{2}V_{2}-P_{1}V_{1}=n_{2}RT-n_{1}RT

=(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^{-4}Scm^{-1}

To find: b = ?

Solution : `k=b/R`

b=kxR

=1.46x10^{-4}x1500

b=2190x10^{-4} cm^{-1}

b = 0.219cm^{-1} |