In the Haber process, #"30 L"# of #"H"_2# and #"30 L"# of #"N"_2# were used for a reaction which yielded only #50%# of the expected product. What will be the composition of gaseous mixture under these condition in the end?

Question

Mateo Aguilar · Accepted Answer

Here's what I got.

Write the balanced chemical equation that characterizes this reaction first, as you should always do. #color(blue)(3)"H"_ (2(g)) + "N"_ (2(g)) -> color(darkgreen)(2)"NH"_( 3(g))# Your first goal here is to figure out how much ammonia would be produced for a #100%# yield. Use the fact that the mole ratio between the species involved in the reaction is equal to a volume ratio when working at constant pressure and temperature to accomplish that. In other words, at #100%# yield, your reaction will consume #color(blue)(3)# liters of hydrogen gas and #1# liter of nitrogen gas and produce #color(darkgreen)(2# liters of ammonia. It should now be clear that you are working with a limiting reagent because you can see that for all of the hydrogen gas to react, you need #30 color(red)(cancel(color(black)("L H"_2))) * "1 L N"_2/(color(blue)(3)color(red)(cancel(color(black)("L H"_2)))) = "10 L N"_2# Since you have more than #"10 L"# of nitrogen gas, you can say that nitrogen is in excess, which implies that hydrogen gas is the limiting reagent here. Now, here's where things get a little tricky. At #100%# yield, the reaction will consume #"30 L H"_2 -># all the hydrogen is consumed because it acts as a limiting reagent #"10 L N"_2"# and generate #30 color(red)(cancel(color(black)("L H"_2))) * (color(darkgreen)(2)color(white)(a)"L NH"_3)/(color(blue)(3) color(red)(cancel(color(black)("L H"_2)))) = "20 L N"_2# This means that after the reaction is complete and at #100%# yield, the reaction vessel would contain However, you know that the reaction has a #50%# yield. This basically means that the reaction will only produce #50%# of the #"20 L"# of nitrogen it would produce at #100%# yield. Put differently, your response generates #1/2 * "20 L NH"_3 = "10 L NH"_3# The volumes of the two reactants that did not ultimately produce nitrogen gas should now be assumed to have remained unreacted. It is essentially assumed that only half of the volumes of the two reactants that were really involved in the reaction were converted to nitrogen gas, with the remaining half remaining unreacted. This means that after the reaction is complete and at #50%# yield, the vessel will contain #"H"_2:" "overbrace(1/2 * "30 L H"_2)^(color(purple)("unreacted")) = "15 L H"_2# #"N"_2: " "overbrace("20 L N"_2)^(color(blue)("in excess")) + overbrace(1/2 * "10 L N"_2)^(color(purple)("unreacted")) = "25 L N"_2# #"NH"_3:" " 1/2 * "20 L NH"_3 = "10 L NH"_3#

Xavier Avery · Answer

undefined In the Haber process, if 30 L of H₂ and 30 L of N₂ are used for a reaction yielding only 50% of the expected product, the composition of the gaseous mixture at the end will consist of unreacted H₂ and N₂, as well as the produced NH₃. The molar ratio of the reactants and products in the balanced equation $N₂ + 3H₂ \rightarrow 2NH₃$ can be used to determine the amounts of each component in the final mixture.