Question

In Haber process, 30 L of dihydrogen and 30 L of
dinitrogen were taken for reaction which yielded only
50% of the expected product. What will be the
composition of the gaseous mixture under the
aforesaid conditions in the end ?
(a) 20 L NH3, 25 L N, and 20 L H,
(6) 10 L NH3, 25 L N, and 15 L H,
(c) 20 L NH3, 10 L N, and 30 L H,
(d) 20 L NH3, 25 L N, and 15 LH,
(C.B.S.E. Med. 2003)
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Answer 1

Answer:

Here's what I got.  

Explanation:

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

3H1(g) + N2(g) → 2NH3(g)  {(g) is Grams}

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  3  liters of hydrogen gas and  1  liter of nitrogen gas and produce  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

30LH2 . 1LN2/3LH2 = 10LN2     (As the 30LH2 get Divided By 3LH2)

[ i.e., 10 . 1LN2 = 10LN2].

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

30

L H

2 . 2LNH3/3LH2 = 20L N2

(Again 30L H2 is Divided By 3LH2)

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
• 20LNH3

Hope u got ur answer