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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 kJ mol−1, respectively. The value of △Uo300 for the reaction:
CCl4(l)+2H2O(g)→CO2(g)+4HCl(g) is
The standard enthalpies of formation at 300 K for CCl4(l), H2O (g), CO2(g) and HCl(g) are −107,−242,−394 and−93 kJ mol−1, respectively. The value of △Uo300 for the reaction:
CCl4(l)+2H2O(g)→CO2(g)+4HCl(g) is
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Solution
The correct option is C −182.5 kJ mol−1
For the reaction CCl4(l)+2H2O(g)→CO2(g)+4HCl(g),)
we have
△rH=Enthalpy of reaction△fH=Enthalpy of formation
△rH=△fH(CO2,g)+4△fH(HCl,g)−△FH(CCl4,l)−2△fH(H2O,g)
Putting all the values, we get
△rH=(−394−4×93+107+2×242) kJ mol−1=−175 kJ mol−1
Also, We know;
△H=△U+△ngRT
where, △ng is number of change in gaseous moles.
△U=△H−△ngRT
△ng=5−2=3
putting all values in the formula,
△U=△H−△ngRT
=−175 kJ−(3×8.314×300×10−3 kJ) =−182.5 kJ mol−1
For the reaction CCl4(l)+2H2O(g)→CO2(g)+4HCl(g),)
we have
△rH=Enthalpy of reaction△fH=Enthalpy of formation
△rH=△fH(CO2,g)+4△fH(HCl,g)−△FH(CCl4,l)−2△fH(H2O,g)
Putting all the values, we get
△rH=(−394−4×93+107+2×242) kJ mol−1=−175 kJ mol−1Also, We know;
△H=△U+△ngRT
where, △ng is number of change in gaseous moles.
△U=△H−△ngRT
△ng=5−2=3
putting all values in the formula,
△U=△H−△ngRT
=−175 kJ−(3×8.314×300×10−3 kJ) =−182.5 kJ mol−1
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