Question

\[{N_2} + 3{H_2} \rightleftharpoons 2N{H_3}\] Starting with one mole of nitrogen and 3 moles of hydrogen, at equilibrium 50% of each had reacted. If the equilibrium pressure is P, the partial pressure of hydrogen at equilibrium would be:
(A) P/2
(B) P/3
(C) P/4
(D) P/6

Answer 1

Hint: Try to recall that partial pressure of a gas is equal to the product of the total pressure and mole fraction of gas and mole fraction is the ratio of number of moles of a substance to total moles. Now, by using this you can easily find the correct option from the given ones.
Complete answer:
* The equilibrium reaction is : \[{N_2} + 3{H_2} \rightleftharpoons 2N{H_3}\]
Initial moles of nitrogen, \[{N_2}\]= 1
Initial moles of hydrogen,\[{H_2}\]=3
We are given that at equilibrium 50% of each reactant had reacted
So, number of moles of nitrogen dissociated, \[{N_2}\]\[{\text{ = }}\dfrac{{{\text{50}}}}{{{\text{100}}}}{\text{ \times 1 = 0}}{\text{.5mole}}\]
Number of moles of hydrogen dissociated, \[{H_2}\]\[{\text{ = }}\dfrac{{{\text{50}}}}{{{\text{100}}}}{\text{ \times 3 = 1}}{\text{.5mole}}\].
Therefore, amount of \[{N_2}\]$, $\[{H_2}\]$ and $\[N{H_3}\] at equilibrium will be
\[{N_2} = 1 - 0.5 = 0.5mole\]
\[{H_2} = 3 - 1.5 = 1.5mole\]
Since, 1 mole of \[{N_2}\] on dissociation gives 2 mole of \[N{H_3}\]
So, 0.5 mole of \[{N_2}\] gives\[2 \times 0.5 = 1mole{\text{ of }}N{H_3}\].
Number of moles of hydrogen left at equilibrium= 1.5 mole
Total number of moles at equilibrium\[ = 0.5 + 1.5 + 1 = 3mole\]
Mole fraction of hydrogen \[ = \dfrac{{number{\text{ of moles of hydrogen}}}}{{total{\text{ number of moles at equilibrium}}}} = \dfrac{{1.5}}{3} = 0.5\]
Also, given total pressure at equilibrium is P
Partial pressure of hydrogen \[ = mole{\text{ fraction of hydrogen}} \times {\text{total pressure at equilibrium}}\]
 \[
    \\
  {\text{ = }}\left( {0.5} \right) \times P \\
  or,\dfrac{P}{2} \\
\]
Therefore, from the above calculation, we can say that option A is the correct option to the given question.
Note:
* It should be remembered to you that the law of mass action states that the rate at which a substance reacts is directly proportional to its active mass and hence the rate of a chemical reaction is directly proportional to the product of the active masses of the reactants.
* Also, you should remember that equilibrium constant of a reaction is constant at constant temperature and does not depend upon the concentration of reactants.