Tag: preparation of ammonia-laboratory method and haber's process

Questions Related to preparation of ammonia-laboratory method and haber's process

Which compounds is related to Haber's process:

chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

  1. $CO _2$

  2. $H _2$

  3. $NO _2$

  4. $NH _3$

Show answer
Correct Option: D
Explanation:

Haber's process is used to synthesize ammonia ($NH _3$) from nitrogen and hydrogen gases using in 1:3 ratio and $Fe,Mo$ is used as catalyst and promoter respectively.
$N _2+3H _2\xrightarrow[Fe]{Mo}2NH _3$
Hence option D is correct

In the Haber process of synthesis of $NH _3$,  which of the following statement is correct?

chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

  1. $Mo$ acts as a catalyst and $Fe$ as a promoter

  2. $Fe$ acts as a catalyst and $Mo$ as a promoter

  3. $Fe$ acts as inhibitor and $Mo$ as a catalyst

  4. $Fe$ acts as promoter and $Mo$ as autocatalyst

Show answer
Correct Option: B
Explanation:

Preparation of Ammonia: 

$N _2 + 3H _2\rightarrow 2NH _3$

At high pressure about 200-250 atm, and a moderate temperature of $750K$, the reaction takes place. Removal of ammonia is done by liquefaction, In the process, iron is used as a catalyst and molybdenum as a promoter.

Which compound of nitrogen is formed when $CaNH$ reacts with hot water?

chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

  1. $NH _{3}$

  2. $N _{2}O$

  3. $NH _{2}NH _{2}$

  4. $NO _{2}$

Show answer
Correct Option: A
Explanation:

When $CaNH$ reacts with hot water it forms $NH _3,Ca(OH) _2$.

$CaNH+2H _2O\rightarrow Ca(OH) _2+NH _3$.

Henceoption  is correct.

Formation of ammonia will be favorable in the following cases(s):
$N _2(g)\,+\,3H _2(g)\rightarrow2NH _3(g)$      $\Delta H\,<\,0$

chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

  1. Addition of inert gas at constant volume

  2. Decreasing the volume of the container

  3. Addition of inert gas at constant pressure

  4. Increasing the pressure

Show answer
Correct Option: D
Explanation:
  • In accordance with Le Chatelier’s principle, high pressure would favour the formation of ammonia. The optimum conditions for the production of ammonia are a pressure of 200 × 105 Pa (about 200 atm), a temperature of ~ 700 K and the use of a catalyst such as iron oxide with small amounts of $K _2O$ and $Al _2O _3$ to increase the rate of attainment of equilibrium.
  • Addition of an inert gas at constant volume: When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change.
  • Hence, option A is correct answer as addition of ideal gas increases pressure and thus favours formation of Ammonia.

Consider the equilibrium $NH _{4}Cl (s)\rightleftharpoons NH _{3}(g) + HCl(g)$. An inert gas is added to the system at constant volume and temperature. The correct statement(s) is/ are:

chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

  1. the partial pressures of $NH _{3}$ and $HCl$ in the system will increase

  2. the partial pressures of $NH _{3}$ and $HCl$ in the system will remain the same

  3. the partial pressures of $NH _{3}$ and $HCl$ in the system will decrease

  4. the entropy of the system will increase

Show answer
Correct Option: B,D
Explanation:

$NH _4$ $+$ $Cl(s)$ $\rightleftharpoons$ $NH _3 (g)$ $+$ $HCl$.

Noble gas are  inert hence adding inert gas doesn't effect the partial pressure.
This is because noble gas gets added to both reactant and product.
$X (g)$ $+$ $NH _4Cl (S)$ $ \rightleftharpoons $ $NH _3(g)$ $+$ $HCl(g)$ $+$ $X(g)$
Where X=Noble gas
Hence the partical presence of $NH _3$ and $HCl$ will remain same.
The amount of gas in the system will increase after adding inert gas.
Since gases has high entropy than liquid and solid, the entropy  of the system will increase.  

Magnesium on heating to redness in an atmosphere of $N _{2}$ and then on treating with $H _{2}O$ gives?

chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

  1. $NH _{3}$

  2. $H _{2}$

  3. $N _{2}$

  4. $O _{2}$

Show answer
Correct Option: A
Explanation:
Megnesium when heated with $N _2$ in atmosphere it forms magnesium nitride($Mg _3N _2$) which on hydrolysis(reaction with water) gives ammonia.
$3Mg+N _2\rightarrow Mg _3N _2$
$Mg _3N _2+6H _2O\rightarrow 3Mg(OH) _2+2NH _3$.
Hence option A is correct.

Ammonia gas can be collected by the displacement of:

chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

  1. Conc. $H _2SO _4$

  2. Brine

  3. Water

  4. Mercury

Show answer
Correct Option: D
Explanation:

Ammonia is less dense than air hence can be collected by upward displacement of air and mercury.

In Haber's process $50.0\ g$ of $N _{2}(g)$ and $10.0\ g$ of $H _{2}(g)$ are mixed to produced $NH _{3}(g)$. What are the number of moles of $NH _{3}(g)$ formed?

chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

  1. $3.33$

  2. $2.36$

  3. $2.01$

  4. $5.36$

Show answer
Correct Option: A
Explanation:
$N _2 + 3H _2 \rightarrow2NH3$
Here.
28 g $N _2$ reacts with 6 g of $H _2$
So,
50 g of $N _2$ react with $\dfrac{6}{28}\times50 g -H _2 = 10.7 g$ but we have only 10g so $H _2$ is limiting reagent..

6 g $H _2$ gives 34g $NH _3$
10g $H _2$ will give $\dfrac{34}{6}\times10=56.67g -NH _3$
So number of moles of $NH _3$ formed is,
$=\dfrac{Given-mass}{Molar-mass}=\dfrac{56.67}{17}=3.33-moles$