For the reaction, heats of formation of N2H4,H2O2,H2O are 12, -45 and -57.8 kcalmol−1. Internal energy change for this reaction at 298K is: N2H4(l)+2H2O2(l)→N2(g)+4H2O(g)
A
- 153.2 kcalmol−1
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B
- 641.142 kJmol−1
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C
-24.8 kcalmol−1
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D
-309 kcalmol−1
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Solution
The correct option is A - 153.2 kcalmol−1 N2H4(l)+2H2O2(l)⟶N2(g)+4H2O(g)ΔHreaction=∑ΔHf(products)−∑ΔHf(reactants)=ΔHf(N2(g))+4ΔHf(H2O(g))−[ΔHf(N2H4(l))+2ΔHf(H2O2(l))]=0+4×(−57.8)−2×(−45)−12=−153.2kcalmol−1Now,ΔH=ΔU+ΔngRT∴ΔUreaction=ΔHreaction−ΔngRT=(−153.2)−2×2×298×10−3=−153.2kcalmol−1