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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, and -93 kJmol1 respectively. The value of ΔU for the reaction
CCl4(l)+2H2O(g)CO2(g)+4HCl(g)
at 300 K is:

A
170 kJmol1
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B
175 kJmol1
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C
182.50 kJmol1
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D
282.50 kJmol1
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Solution

The correct option is C 182.50 kJmol1
The reaction given is,
CCl4(l)+2H2O(g)CO2(g)+4HCl(g)
So, ΔH=ΔHf(CO2,g)+4ΔHf(HCl,g)ΔHf(CCl4,l)2ΔHf(H2O,g)
ΔH=(3944×93+107+2×242) kJmol1
ΔH=175 kJmol1
We know,
ΔH=ΔU+ΔngRT
ΔU=ΔHΔngRT
=175(3)(8.314×103 kJK1mol1)(300 K)182.48 kJmol1

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