Question

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?

Answer 1

Here's what I got.

Explanation:

As always, start by writing the balanced chemical equation that describes this reaction

`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.

To do that, use the fact that when working with constant pressure and temperature, the mole ratio that exists between the species involved in the reaction is equivalent to a volume ratio.

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.

At this point, it should be obvious that you're dealing with a limiting reagent. Notice that in order for all 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 produce

`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

• `"0 L H"_2 ->` completely consumed
• `"30 L N"_2 - "10 L N"_2 = "20 L N"_2 ->` in excess
• `"20 L NH"_3`

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.

In other words, your reaction produces

`1/2 * "20 L NH"_3 = "10 L NH"_3`

At this point, you should assume that the volumes of the two reactants that did not end up producing nitrogen gas remained unreacted.

Basically, you are to assume that only half of the volumes of the two reactants that actually took part in the reaction ended up producing nitrogen gas. The other half are left 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`