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Question

The standard enthalpies of formation at 300 K for CCl4(l), H2O (g), CO2(g) and HCl(g) are 107,242,394 and93 kJ mol1, respectively. The value of Uo300 for the reaction:

CCl4(l)+2H2O(g)CO2(g)+4HCl(g) is
x kJ mol1. Find the value of x
(Take R=8.314 JK1mol1)

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Solution

For the reaction CCl4(l)+2H2O(g)CO2(g)+4HCl(g),)
we have
rH=Enthalpy of reactionfH=Enthalpy of formation
rH=fH(CO2,g)+4fH(HCl,g)FH(CCl4,l)2fH(H2O,g)

Putting all the values, We get;

rH=(3944×93+107+2×242) kJ mol1=175 kJ mol1
Also, We know;
H=U+ngRT
where, ng is number of change in gaseous moles.
U=HngRT
ng=52=3
putting this and other values in the formula above,
U=HngRT
=175 kJ(3×8.3×300×103 kJ) =182.47 kJ mol1

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