Question

The density of an equilibrium mixture of N2O4 and NO2 at 1 atm and 348 K is 1.84 g/L . Calculate Kc and Kp for the equilibrium.
Can I take initial moles as 1 in these kind of problems ?

Answer 1

Answer : The $K_p$ and $K_c$ is, 5.17 atm and 0.1811 mole/L

Solution :

First we have to calculate the molar mass of the mixture.

formula used : $M=\frac{dRT}{P}$

where,

M = molar mass of the mixture

d = density of the mixture = 1.84 g/L

R = gas constant = 0.082 Latm/moleK

T = temperature of the mixture = 348 K

P = pressure of the mixture = 1atm

Now put all the given values in this formula, we get the molar mass of the mixture.

$M=\frac{(1.84g/L)\times (0.082Latm/moleK)\times (348K)}{1atm}=52.50g/mole$

The equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Let the mole fraction of $N_2O_4$ is 'x' and the mole fraction $NO_2$ is 'y'.

Now we have to calculate the mole fractions of $N_2O_4$ and $NO_2$.

As we know, $\bar{M}=\sum x_iM_i$

where, $x_i$ is mole fraction and $M_i$ is molar mass.

By using this expression, two equations will be formed i.e

$52.50=92x+46y\\\\x+y=1$

By solving these two equation, we get the value of 'x' and 'y'.

x = 0.142  and  y = 0.857

Now we have to calculate the value of $K_p$ at total pressure 1 atm.

$K_p=\frac{(x_{NO_2}P)^2}{x_{N_2O_4}P}=\frac{(x_{NO_2})^2}{x_{N_2O_4}}=\frac{(0.857)^2}{0.142}=5.17atm$

Now we have to calculate the value of $K_c$.

$K_p=K_c[RT]^{\Delta n}$

$\Delta n$ = 2 - 1 = 1 moles

Now put the values in this relation, we get

$5.17atm=K_c[(0.082Latm/moleK)\times (348K)]^{1}$

$K_c=0.1811mole/L$

Therefore, the $K_p$ and $K_c$ is, 5.17 atm and 0.1811 mole/L

Answer 2

Explanation:

First we have to calculate the molar mass of the mixture.

formula used : M=\frac{dRT}{P}M=

P

dRT

where,

M = molar mass of the mixture

d = density of the mixture = 1.84 g/L

R = gas constant = 0.082 Latm/moleK

T = temperature of the mixture = 348 K

P = pressure of the mixture = 1atm

Now put all the given values in this formula, we get the molar mass of the mixture.

M=\frac{(1.84g/L)\times (0.082Latm/moleK)\times (348K)}{1atm}=52.50g/moleM=

1atm

(1.84g/L)×(0.082Latm/moleK)×(348K)

=52.50g/mole

The equilibrium reaction is,

N_2O_4(g)leftharpoons 2NO_2(g)N

2

O

4

(g)leftharpoons2NO

2

(g)

Let the mole fraction of N_2O_4N

2

O

4

is 'x' and the mole fraction NO_2NO

2

is 'y'.

Now we have to calculate the mole fractions of N_2O_4N

2

O

4

and NO_2NO

2

.

As we know, \bar{M}=\sum x_iM_i

M

ˉ

=∑x

i

M

i

where, x_ix

i

is mole fraction and M_iM

i

is molar mass.

By using this expression, two equations will be formed i.e

\begin{gathered}52.50=92x+46y\\\\x+y=1\end{gathered}

52.50=92x+46y

x+y=1

By solving these two equation, we get the value of 'x' and 'y'.

x = 0.142 and y = 0.857

Now we have to calculate the value of K_pK

p

at total pressure 1 atm.

K_p=\frac{(x_{NO_2}P)^2}{x_{N_2O_4}P}=\frac{(x_{NO_2})^2}{x_{N_2O_4}}=\frac{(0.857)^2}{0.142}=5.17atmK

p

=

x

N

2

O

4

P

(x

NO

2

P)

2

=

x

N

2

O

4

(x

NO

2

)

2

=

0.142

(0.857)

2

=5.17atm

Now we have to calculate the value of K_cK

c

.

K_p=K_c[RT]^{\Delta n}K

p

=K

c

[RT]

Δn

\Delta nΔn = 2 - 1 = 1 moles

Now put the values in this relation, we get

5.17atm=K_c[(0.082Latm/moleK)\times (348K)]^{1}5.17atm=K

c

[(0.082Latm/moleK)×(348K)]

1

K_c=0.1811mole/LK

c

=0.1811mole/L

Therefore, the K_pK

p

and K_cK

c

is, 5.17 atm and 0.1811 mole/L