The standard enthalpies of formation at 300 K for CCl4(l),H2O(g),CO2(g)andHCl(g) are -107, -242, -394, and -93 kJmol−1 respectively. The value of ΔU for the reaction CCl4(l)+2H2O(g)→CO2(g)+4HCl(g) at 300 K is:
A
−170kJmol−1
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B
−175kJmol−1
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C
−182.50kJmol−1
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D
−282.50kJmol−1
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Solution
The correct option is C−182.50kJmol−1 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=(−394−4×93+107+2×242)kJmol−1 ⇒ΔH=−175kJmol−1 We know, ΔH=ΔU+ΔngRT ⇒ΔU=ΔH−ΔngRT =−175−(3)(8.314×10−3kJK−1mol−1)(300K)≈−182.48kJmol−1