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Q.

Consider the following reaction OF2g+H2Og→O2g+2HFg if Hf0 for OF2g,H2Og and  HFgare +20,-250,-270 KJmol−1at 298K then calculate the standard internal energy change for the reaction

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a

-312.47KJ

b

+312.47KJ

c

-213.47KJ

d

+213.47KJ

answer is A.

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Detailed Solution

ΔHreaction 0=∑HP0-∑HR0ΔH0=2×HHF0+HO20-HOF20+HH2O0=[2×(-270)-0]-[20-250]∵HO20=0=-310KJΔH0=ΔU0+Δ∩RTΔn=3-2=1;R=8.314Jk-1 mol-1,T=298 K∴-310×103=ΔU0+1×8.314×298ΔQ0=-312477.5 J=-312.47KJ
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