Calculate degree of dissociation

[utkarshakash]

Homework Statement

Vapour density of $N_2O_4$ which dissociated according to the equation
$N_2O_4(g) \Leftrightarrow 2NO_2(g)$ is 25.67 at 100°C and a pressure of 1 atm. Calculate the degree of dissociation and $K_p$ for the reaction

Homework Equations

The Attempt at a Solution

$\alpha = \frac{D_t-D_o}{D_o}$

$D_t = \frac{92}{2}$
Plugging in the values does not yield desired answer.

 

[Borek]

Please elaborate on what you are doing. What are D_t and D₀? What is 92?

Is that the whole question? Either I am missing something or it can't be solved as given.

 

[Saitama]

Please elaborate on what you are doing. What are D_t and D₀? What is 92?

Is that the whole question? Either I am missing something or it can't be solved as given.

I suppose the OP directly posted the equations without showing how he/she has got it. If i remember correctly, Dt and Do are the theoretical and observed vapour density of a mixture at equilibrium. 92 is the molar mass of N2O4.

 

[utkarshakash]

I suppose the OP directly posted the equations without showing how he/she has got it. If i remember correctly, Dt and Do are the theoretical and observed vapour density of a mixture at equilibrium. 92 is the molar mass of N2O4.

Yes that's right. But the answer which I'm getting is wrong.

 

[DeeNos]

First, let's define some terms. The degree of dissociation, denoted by alpha (α), is defined as the fraction of molecules that have dissociated in a reaction. In this case, it represents the fraction of N2O4 molecules that have dissociated into 2NO2 molecules. The vapour density (D) is defined as the mass of a gas in a given volume compared to the mass of an equal volume of air at the same temperature and pressure. It is related to the molar mass (M) of a gas by the equation D = M/RT, where R is the gas constant and T is the temperature in Kelvin.

To calculate the degree of dissociation, we can use the equation given in the problem:
α = (D_t - D_o)/D_o
where D_t is the total vapour density of the mixture and D_o is the initial vapour density of N2O4. However, the values given in the problem are not the vapour densities, but rather the vapour density ratios (D_t/D_o). To find the vapour densities, we can use the equation D = M/RT and the given temperature and pressure. For N2O4, the molar mass is 92 g/mol. Plugging in these values, we get:
D_t = (92 g/mol)/(0.0821 L*atm/mol*K * 373 K) = 2.623 g/L
D_o = (92 g/mol)/(0.0821 L*atm/mol*K * 373 K) = 2.623 g/L
Note that the initial and total vapour densities are the same, since we are starting with pure N2O4.

Plugging these values into the equation for alpha, we get:
α = (2.623 g/L - 2.623 g/L)/2.623 g/L = 0
This means that at 100°C and 1 atm, there is no dissociation of N2O4 into 2NO2. This makes sense, since the reaction is endothermic and would require energy to break the bonds in N2O4.

To calculate the equilibrium constant (Kp) for the reaction, we can use the equation:
Kp = (P_NO2)^2/P_N2O4
where P_NO2 and P_N2O4 are the partial pressures of NO2 and N2