Question

0.6 mole of $N{{H}_{3}}$ in a reaction vessel of 2$d{{m}^{3}}$ capacity was brought to equilibrium . The vessel was then found to contain 0.15 mole of${{H}_{2}}$ formed by the reaction $2N{{H}_{3}}(g){{N}_{2}}(g)+3{{H}_{2}}(g)$
Which of the following statements is false?
A. 0.15 mole of the original $N{{H}_{3}}$ had dissociated at equilibrium
B. 0.55 mole of ammonia is left in the vessel
C. At equilibrium the vessel contained 0.45 mole of ${{N}_{2}}$
D. The concentration of $N{{H}_{3}}$ at equilibrium is 0.25 mole per $d{{m}^{3}}$

Answer 1

Hint: The process of conversion of ammonia into nitrogen and hydrogen takes place under high pressure and high temperature like 200 atm pressure and nearly 700 K temperature in the presence of catalyst like $A{{l}_{2}}{{O}_{3}}\And {{K}_{2}}O$ .This process is commonly called as Haber’s process in commercial uses.The equilibrium constant of the reaction will be product of the concentration of the products to the reactants before and after the change from these two equations and taking a variable for the concentration of the reactants and product side we can calculate for the conc.

Complete Step by Step Answer:
The capacity of the reaction vessel is of 2$d{{m}^{3}}$ which is given , now 0.6 mol of the reactant which is ammonia is the initial concentration of the reactant. The reaction can be illustrated as:
$2N{{H}_{3}}(g){{N}_{2}}(g)+3{{H}_{2}}(g)$

At this time , the concentration of products side will be 0 mole each for nitrogen and hydrogen. Now let us suppose that after a certain time interval the concentration of the reactant side decreases by a value of x and so this amount of the products will be formed in the product side for sure . Multiplying with the stoichiometric coeff. Of the reactants and the products as per the convenience of the mole of the reactants and products which is to be taken in consideration we will observe that the new concentration of the ammonia will be (0.6-2x) as 2x amount of its amount is dissociated in the reaction and so the in product side now the concentration of ${{N}_{2}}$ will be x mole and that of ${{H}_{2}}$ will be 3x mole . Since , it is given in the question that the amount of hydrogen produced in the reaction process is 0.15 mole , that means –
3x=0.15
X=$\frac{0.15}{3}=0.05$
X=0.05mole
Thus the concentration of $N{{H}_{3}}$ can be obtained at this time as –
[$N{{H}_{3}}$]=0.6-2x=0.6-2×0.05=0.6-0.1=0.5 mole
Now since the capacity of the reaction vessel is 2 $d{{m}^{3}}$ so the concentration of $N{{H}_{3}}$ per $d{{m}^{3}}$ can be obtained as –
[$N{{H}_{3}}$]=0.5mole/2$d{{m}^{3}}$=0.25mol/$d{{m}^{3}}$
Thus , the correct option will be D.

Note: The final concentration is asked in per $d{{m}^{3}}$ this should be taken in mind and so it needs to be divided by 2 in the last process . Many students finalise the answer as 0.5 only which is wrong.