Question

At 340K and 1 atm pressure, ${N}_{2}{O}_{4}$ is 66% dissociated into $N{O}_{2}$. What volume of 10g ${N}_{2}{O}_{4}$ occupy under these conditions?
A. V=10 L
B. V=2.5 L
C. V=5.04 L
D. V=1.5 L

Answer 1

Hint: Dissociation of a compound refers to the breaking of the bigger molecules into small particles, ions or radicals. And the conditions in which the dissociation takes place is also very important.

Complete step by step answer: Let us first look at the reaction involved in the above question.
$\underset { Dinitrogen\quad Tetroxide }{ { N }_{ 2 }{ O }_{ 4 } } \quad \rightleftharpoons \quad \underset { Nitrogen\quad oxide }{ N{ O }_{ 2 } }$

Now, it is given that only 66% of ${N}_{2}{O}_{4}$ dissociated into $N{O}_{2}$. Therefore, the dissociation constant, $\alpha$ = 0.66.

Now, let us assume that at time, t=0, the number of moles of ${N}_{2}{O}_{4}$ and $N{O}_{2}$ are 1 and 0 respectively. Then at time, $t={t}_{eq}$, the number of moles of ${N}_{2}{O}_{4}$ will be $1 - \alpha$ and the number of moles of $N{O}_{2}$ will be $2\alpha$.

| ${N}_{2}{O}_{4}$| $N{O}_{2}$
---|---|---
No. of moles at t=0| 1| 0
No. of moles at $t={t}_{eq}$| $1- \alpha$| $2 \alpha$

Therefore, the total number of moles at equilibrium= $1 - \alpha + 2\alpha$
Total number of moles = $1 + \alpha$

Now, we know that the value of $\alpha$ is 0.66. Substituting $\alpha$ in the total number of moles, we get
Total number of moles = 1 + 0.66 = 1.66 moles.

Now, we need to find the no. of moles in 10g of ${N}_{2}{O}_{4}$
Therefore, $No.\quad of\quad moles\quad in\quad 10g\quad { N }_{ 2 }{ O }_{ 4 }\quad =\quad \cfrac { Weight }{ Molecular\quad mass }$
The molecular mass of ${N}_{2}{O}_{4}$ is 92g.
$No.\quad of\quad moles\quad in\quad 10g\quad { N }_{ 2 }{ O }_{ 4 }\quad =\quad \cfrac { 10 }{ 92 } \quad moles$
Now, at equilibrium, if 1 mole of ${N}_{2}{O}_{4}$ dissociates to form 1.66 total moles, then $\cfrac {10}{92}$ moles of ${N}_{2}{O}_{4}$ will dissociate to give $\quad \cfrac { 10 }{ 92 } \quad \times \quad 1.66\quad =\quad 0.18\quad moles$

Now, we know that ideal gas law is given as follows
$PV = nRT$
where P is pressure, V is volume, n is total no. of moles, T is temperature and R is gas constant.
Solving for volume, V, we get
$V\quad =\quad \cfrac { nRT }{ P }$
Here, we are given from the question that P=1 atm, T=340K, R=0.0821 L atm/mol K and we calculated above that n=0.18 moles. Substituting these value in the above equation, we get
$V\quad =\quad \cfrac { 0.18\quad \times \quad 0.082\quad \times \quad 340 }{ 1 }$
$\implies V\quad =\quad 5.04\quad L$

Therefore, the volume of 10g ${N}_{2}{O}_{4}$ occupied is 5.04 L. Hence, the correct answer is option (C).

Note: While putting the value of the gas constant, R, do make sure that you are using the correct value with the correct units. There are different values of R in different units, like 8.3145 J/mol K, $8.2057 {m}^{3} atm/mol K$, 0.0821 L atm/mol K, etc.