Equilibria Kc question 9

In the equilibrium:

\dpi{100} N_2O_4\rightleftharpoons 2NO_2

2.50g of dinitrogen tetroxide occupy a volume of 1.00dm3 at 1atm and 25oC. Calculate: a) the degree of dissociation, and b) the equilibrium constant

When reading this question it is important to note that the 2.50g are not just for the dinitrogen tetraoxide, but rather an average of all the gases present, since the N2O4 would decompose.

In order to start working this question it is important to find the number of moles present at equilibrium, and this can be done using Pv=nRT.

P would be equivalent to 1 atm, or 101325Pa, volume is 1 dm3 or 1*10-3 m3 and the Temperature would be 25oC or 298K.

\dpi{100} n=\frac{PV}{RT}=\frac{101325*1*10^-^3}{8.314*298}

Where the number of moles is found to be 0.04090 moles.

Using the values for the moles obtained as the total moles, moles of  N2O4 as x and NO2 as y, it is possible to find the individual components by calculating the mass of each component.

\dpi{100} RMM_{N_2O_4}*moles_{N_2O_4}+RMM_{NO_2}*moles_{NO_2}=total\;mass
\dpi{100} 92x+46y=2.5g

Since x+y=total moles, then it can be said that y = 0.04090 – x, which can be substituted in the above equations.

\dpi{100} 92x+46(0.04090-x)=2.5g

where x can be found to be 0.01348 moles.

This can be used to note that N2O4 has 0.01345 moles and NO2 has 0.02745 moles.

An ICE table can then be used in order to be able to calculate the dissociation constant and kp.

Since 0.02745 moles of NO2 are present then it is possible to find that c is \dpi{100} \frac{0.02745}{2} which is equivalent to 0.01373moles.

This can then be used to find the original moles of N2O4 which would be equivalent to 0.01345+0.01373 which is equal to 0.02718 moles.

The dissociation constant can then be found to be:

\dpi{100} dissociation\; constant =\frac{moles\; dissociated}{total\;moles}=\frac{0.01373}{0.02718}

The dissociation constant is therefore 0.05052.

The kc equation is: \dpi{100} k_c=\frac{[NO_2]^2}{[N_2O_4]}

Since the volume is 1L, then the concentration is the same as the number of moles. Once the concentrations are found, then it is just a matter of placing each value in the kc equation as follows:

\dpi{100} k_c=\frac{(0.02745)^2}{0.01373}

where kc would be found to be: 0.0549 mol dm-3