Question

Consider the dissociation reaction:N2O4(g) ? 2 NO2 (g)The DGrxn?for this reaction at 55 ?C is ?0.8385 kJ/mol. The density of an equilibrium mixture of NO2and N2O4gases is found to be 5.12 g/L at 55 ?C and a certain pressure. Assuming these gases behave ideally, calculate:(a)

The degree of dissociation of N2O4.(b)

The average molecular weight of the mixture.(c)

The total pressure of the gas mixture.

Answer 1

The free enthalpy change $\Delta G_{rxn} =- 0.8385 \ kJ/mol$ .

Temperature = $T = 55 \ ^\circ C = 55 + 273 \ K = 328 \ K$

R, gas constant =

Now, $\Delta \ G_{rxn}$ and $K_{eq}$ are related as follows:

$\Delta G_{rxn} = -RTln K_{eq}$

Hence,

$K_{eq} = e^{\frac{-\Delta G_{rxn}}{RT}}\\ = e^{\frac{-(-838.5 \ Jmol^{-1})}{8.314 \ J mol^{-1} K^{-1} \times 328 \ K}} \\ =e^{0.307} = 1.36$

$K_{eq}$ is also given by the following equation for our decomposition of $N_2O_4_{(g)} \rightarrow 2NO_2_{(g)}$

$K_{eq} = \frac{[NO_2]_{eq}^2}{[N_2O_4]_{eq}}$

a)

Now we can prepare a table to follow the concentration change of reaction and product through the course of the reaction.

$N_2O_4_{(g)} \rightarrow 2NO_2_{(g)}$

	$[N_2O_4]$	$[NO_2]$
Initial Concentration, mol/L	1	0
Change	-x	+2x
Equilibrium Concentration, mol/L	1-x	2x

Hence,

$K_{eq} = \frac{[NO_2]_{eq}^2}{[N_2O_4]_{eq}} =1.36 \\ \Rightarrow \frac{(2x)^2}{1-x} =1.36 \\ \Rightarrow 4x^2 +1.36 x -1.36 = 0 \\ \Rightarrow x = 0.437 \ or \ x = -0.777$

Since, x is a concentration, it cannot be negative. Hence x = 0.437.

Hence, the degree of dissociation or the fraction of moles of $N_2O_4$ that underwent dissociation is x = 0.437.

b)

Average molecular weight = mole fraction of $N_2O_4$ * molar mass of $N_2O_4$ + mole fraction of $NO_2$ * molar mass of $NO_2$

Molar mass of $NO_2$ = 46 gm/mol

Mole fraction of $NO_2$ in equilibrium mixture is

$\frac{2x}{1-x+2x} = \frac{2x}{1+x} = \frac{2\times 0.437}{1+0.437} = \frac{0.874}{1.437} = 0.608$

Molar mass of $N_2O_4$ = 92 gm/mol

Mole fraction of $N_2O_4$ in equilibrium mixture is

$\frac{1-x}{1-x+2x} = \frac{1-x}{1+x} = \frac{1-0.437}{1+0.437} = \frac{0.563}{1.437} = 0.392$

Hence, average molecular weight is

$0.392 \times 92 \ gm/mol + 0.608 \times 46 \ gm/mol \\=36.064 \ gm/ mol + 27.968 \ gm/mol = 64.032 \ gm/mol$

Hence, the average molecular weight of the reaction mixture is 64.032 gm/mol.

c)  Since the gas obeys ideal gas laws

for the mixture of gas.

$p =\frac{nRT}{V}$

Now substituting   $n =\frac{ m}{Mm}$

Where m = mass of the gas mixture and Mm is molar mass of the gas mixture.

$p = \frac{m}{V} \times \frac{RT}{Mm}$

Replace $\frac{m}{V} = \rho$ (density).

$p = \rho\times \frac{RT}{Mm}$

We know,

$\rho = 5.12\ gm/L \\ Mm = 60.032\ gm/mol \\ R = 0.082 \ L \ atm \ mol^{-1} K^{-1} \\ T = 328 \ K$

Hence,

total pressure of the gas is

$p = 5.12 \ gm/L \times \frac{0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 328 \ K }{64.032 \ gm/mol} = 2.15 \ atm$

hence, the total pressure of the gas is 2.15 atm.