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Given, NH_(3)(g) + 3Cl_(2)(g) NCl_(3)(g) + 3HCl, -DeltaH_(1) N_(2)(g) + Cl_(2)(g) to 2NH_(3)(g),-DeltaH_(2) H_(2)(g) + Cl_(2)(g) to 2HCl_(g),-DeltaH_(3) The heat of formation of NCl_(3)(g) in the terms of DeltaH_(1), DeltaH_(2) and DeltaH_(3) is |
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Answer» `Delta_(r)H = -DeltaH_(1) + (DeltaH_(2))/2 -3/2 DeltaH_(3)` |
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