At 1000 K from the data, N2(g)+3H2(g)→2NH3(g) ΔH∘=−123.77kmol−1Substance N2 H2 NH3Cp/R 3.5 3.5 4Calculate the heat of formation of NH3 at 300 K.
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answer is 44.42 KJ MOL.
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Detailed Solution
Using Kirchhoffs equation:ΔH2(1000K)=ΔH1(300K)+ΔCp(1000−300)ΔCp=2CpNH3−CpN2−3CpH2=−6R=−6×8.314×103kJ∴−123.77=ΔH1(300K)−6×8.314×103×700 or, ΔH1(300K)=−123.77+34.92=88.85kJ for 2mol of NH3ΔHfNH3=−44.42kJmol−1