Question

11 moles \[{N_2}\] and 12 moles of \[{H_2}\] mixture reacted in 20 litres vessel at 800 K. After equilibrium was reached, 6 moles of \[{H_2}\] present. 3.58 litre of liquid water is injected in equilibrium mixture and resultant gaseous mixture suddenly cooled to 300 K. What is the final pressure of gaseous mixture? Negative vapour pressure of liquid solution. Assume (i) all \[N{H_3}\] dissolved in water (ii) no change in volume of liquid (iii) no reaction of \[{N_2}\] and \[{H_2}\] at 300 K.
A. 18.47 atm
B. 60 atm
C. 22.5 atm
D. 45 atm

Answer 1

Hint: This answer will need the concept of stoichiometry of mole concepts, concept of equilibrium and concept of ideal gas equation.
Firstly if we know the final number of moles at equilibrium, Volume of gas occupied, temperature is given in question, then using the ideal gas equation, we can find out easily the pressure of gas at that point.

Complete step by step answer:
We have been given initially 11 moles \[{N_2}\] and 12 moles of \[{H_2}\] mixture, this will result in formation of Ammonia, $N{H_3}$ as the following reaction given below:
\[{N_2} + 3{H_2} \to 2N{H_3}\]
Initial:
\[{\text{ 11 12 0}}\]
At equilibrium, it is given that 6 moles \[{H_2}\] is present.
The amount of \[{H_2}\] used will be total subtracted from remaining at equilibrium.
\[ {\text{Amount of }}{H_2}{\text{ used or reacted = 12}} - 6 = 6moles\]
Using unitary method, we know from equation, 3 moles \[{H_2}\] will produce 2 moles $N{H_3}$ , so 6 moles \[{H_2}\] will produce 4 moles $N{H_3}$.
Now we also know from equation, that 3 moles \[{H_2}\] will need 1 mole of \[{N_2}\] to react, so 6 moles of \[{H_2}\] at equilibrium will react with 2 moles of \[{N_2}\].
Remaining \[{N_2}\] will be a subtraction of total from used.
\[{N_2}{\text{ remaining at equilibrium = }}11 - 2 = 9moles\]
Now, we also know from the given assumption, that the ammonia produced is dissolved in water, so we don’t have any ammonia in gases. The reaction can be written as:
\[N{H_3} + {H_2}O \to N{H_4}OH\]
So the total number of moles at equilibrium will be the sum of moles of \[{N_2}\] and \[{H_2}\] at equilibrium.
\[\therefore {\text{Total moles (n) = moles of }}{{\text{N}}_{\text{2}}}{\text{ + moles of }}{{\text{H}}_{\text{2}}}\]
Substituting the values we get total moles as:
\[n = 6 + 9\]
\[ n = 15moles\]
Now, we need to find the volume of gas occupied. The total volume of container is 20 litres, but it is written in question that 3.58 litres of water is injected, so that volume is occupied by liquid, so the remaining volume of container will be occupied by gas.
\[{\text{Volume of gas occupied = Volume of container }} - {\text{ Volume of liquid}}\]
\[V = 20 - 3.58\]
\[ V = 16.42Litres\]
\[T = 300K\] (given in question)
\[R = 0.0821atm.Lmo{l^{ - 1}}{K^{ - 1}}\] (we have chosen this value as Volume is in litres and we have to get Pressure in atm).
We know ideal gas equation as:
\[PV = nRT\]
Rearrange and get value of pressure as P on one side:
\[P = \dfrac{{nRT}}{V}\]
Now, substitute the above values into the ideal gas equation:
\[P = \dfrac{{15 \times 0.0821 \times 300}}{{16.42}}\]
On simplification and getting the final value of Pressure, we can write as:
\[\therefore P = 22.5atm\]

So, the correct answer is Option C.

Note: We may consider moles of ammonia, $N{H_3}$ into total moles, but remember the assumption given that the ammonia will dissolve in water.
Next common mistake can be for volume, we need to remember that in the ideal gas equation, we have to substitute the volume of gas occupied, not the volume of liquid present, so we should be careful in substituting values of Volume in ideal gas equation.