Question

Ammonium carbamate when heated to $200{}^\circ C$ gives a mixture of \[N{{H}_{3}}\] ​ and \[C{{O}_{2}}\] ​ vapour with a density of 13. What is the degree of dissociation of ammonium carbamate?
A. \[\dfrac{3}{2}\]
B. $\dfrac{3}{4}$
C. $2$
D. $1$

Answer 1

Hint: The percentage of an electrolyte's total moles that disintegrate into its ions when equilibrium is reached is referred to as the electrolyte's degree of dissociation. It’s denoted by $\alpha $.

Formula used: $D=\dfrac{m.wt.}{2}$ where D is the density of ammonium carbamate and m.wt. is the molecular mass of ammonium carbamate. $\alpha =\dfrac{D-d}{(n-1)d}$ where D is the density of ammonium carbamate and d is the given vapour density. n is the number of products.

Complete Step by Step Solution:
Degree of dissociation is defined as the percentage of an electrolyte’s total moles that disintegrates into its ions when equilibrium is reached.

According to given question, ammonium carbamate is disintegrated to give a mixture of \[N{{H}_{3}}\] ​ and \[C{{O}_{2}}\].
Molecular formula of Ammonium carbamate is $N{{H}_{2}}COON{{H}_{4}}$
Molecular mass of Ammonium carbamate= 78
When it’s heated at equilibrium the reaction is given as:
$N{{H}_{2}}COON{{H}_{{{4}_{(s)}}}}2N{{H}_{{{3}_{(g)}}}}+C{{O}_{{{2}_{(g)}}}}$

Now using the formula:
$D=\dfrac{m.wt.}{2}$, putting the values of m.wt= 78
Therefore, $D=\dfrac{78}{2}=39$
Now, we need to find $\alpha $using the formula $\alpha =\dfrac{D-d}{(n-1)d}$

We can see from the equilibrium reaction that $n=3$ and $d=13$, which is given in the question.
Now putting the appropriate values:
$\alpha =\dfrac{39-13}{(3-1)\times 13}$
Solving it further, we get
$\alpha =\dfrac{26}{26}=1$
Hence, the correct option is D. $1$

Note: You should know the basics of equilibrium and how to write and balance a chemical reaction at equilibrium. You must remember the basics of degree of dissociation and the formulas related to it. The different formulas used in the solution must be remembered.