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For the reaction : N_(2)(g)+3H_(2)(g) to 2NH_(3)(g),DeltaH=-24KCal " at " 427^(@)C and 200 atm. Calculate magnitude of internal energy change ( in Kcal DeltaU), if 168 gm N_(2) gas and 30 gm H_(2) gas are allowed to react completely (100% reaction yield ) to form NH_(3) gas at 427^(@)C and 200 atm. |
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Answer» Solution :[0106] MOLES of `N_(2)=(168)/(82)=6"" `Mole of `H_(2)=(30)/(2)=15` Limiting reagent is `H_(2)` `DeltaU=DeltaH=Deltan_(g)RT` `=(-24)xx5-(-2)xx(2)/(1000)xx700xx5=-106 kcal ` |
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