4
$\begingroup$

This is a question I solved in class:

At 35 degrees, $\ce{N_2O_4}$ is 15% dissociated in a flask at a pressure of 1.2 atm. Calculate $K_p$ Equation: $$\ce{N_2O_4(g)<=>2NO_2 (g)}$$

This was how the answer went...

Partial pressure of $\ce{N_2O_4} = 0.85\times 1.2$

Partial pressure of $\ce{NO_2} = 0.1\times 1.2$

then $K_p$ was calculated using the standard formula

My question:

The part I cannot agree to with my teacher is that how is the partial pressure of $NO_2$ equal to 0.15x1.2. If I begin with 1 mole of $\ce{N_2O_4}$, if 15% dissociated at 1.2 atm, I would end up with 0.3 moles of $\ce{NO_2}$... the the partial pressure would be ${0.3\over 1.15}$x$1.2$... this does not go with the general answer stated above

$\endgroup$
  • 2
    $\begingroup$ I would say, you are right: the partial pressure of $\ce{N2O4}$ should be $\frac{0.85}{1.15} \cdot 1.2 \, \text{atm}$ and the partial pressure of $\ce{NO2}$ should be $\frac{0.3}{1.15} \cdot 1.2 \, \text{atm}$. $\endgroup$ – Philipp Mar 8 '14 at 17:29
2
$\begingroup$

I do agree with your point and Philipp explanation. Getting more formal, your $K_p$ at 35°C is:

$$ K^\circ_p(308.15\,\mathrm{K}) = \frac{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)^2\chi^2_\mathrm{NO_2}}{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)\chi_\mathrm{N_2O_4}} $$

Where $p_\mathrm{tot}$ is the pressure in your flask and $p_\circ$ the standard pressure. The $\chi$ quantities stand for molar ratio. For the reaction given, those ratios are:

$$ \chi_\mathrm{NO_2} = \frac{n_{NO_2,0} + 2\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$

and:

$$ \chi_\mathrm{N_2O_4} = \frac{n_{N_2O_4,0} -\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$

Those identities come from amount balance, where $\xi$ is the molar coordinate of the reaction and zero-indexed quantities amount of matter when starting experience.

Assuming that initially there was only $\mathrm{N_2O_4}$, and you are using the following definition of dissociated ratio:

$$ \tau = \frac{\xi}{n_\mathrm{N_2O_4},0} = 0.15$$

You will obtain by substitution, molar ratio values:

$$ \chi_\mathrm{N_2O_4} = \frac{0.85}{1.15}, \chi_\mathrm{NO_2} = \frac{0.3}{1.15}$$

Then you inject those values in the $K^\circ_p$ expression above. Do not forget to express pressures in bars - this is a common mistake, you must remeber that $K_p$ are dimensionless. Therefore, $p_\mathrm{tot} = 1.2 \times 1.01325 \,\mathrm{bar}$ and $p_\circ = 1.0 \, \mathrm{bar}$. This is also an approximation because we are assimilating pressure as activities. But in your case you can consider fugacities close to unity and assimilate your gas mixture as an ideal gas.

Yes, you were right about molar ratio. Do not forget that equilibrium constants are dimensionless: they are ratio of activities!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.