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Suppose we are getting the necessary $\ce{Na}$ for $\ce{HCl}$ out of $\ce{NaCl}$. We need to end up with $V = 100~\mathrm{m^3}$ of $\ce{HCl}$ at standard conditions. Assuming that $\ce{HCl}$ is an ideal gas, and using the ideal gas law, I found there are $$ n = 4272~\mathrm{mol}$$ of matter in those $100~\mathrm{m^3}$. The applied voltage to the cell is $$U = 3.6~\mathrm{V}$$ and the current utilization factor is $\mu=88\%$. Then we also have Faraday's law of electrolysis: $$\frac{m}{M}=\frac{q}{F}$$(the oxidation state of chlorine is $z=1$). Since there is a constant voltage being applied to the cell, the energy used in this reaction is $$E = \frac{UI \Delta t}{0.88} = \frac{Uq}{0.88} = 470~\mathrm{kWh}$$ However the correct answer is $$E=48.9~\mathrm{kWh}.$$ Where did I go wrong? Perhaps it's something blatantly obvious.

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  • $\begingroup$ How did you calculate the number of moles of HCl gas? $\endgroup$ – MaxW Sep 6 '16 at 13:10
  • $\begingroup$ I used the equation $pV=nRT$ $\endgroup$ – Emir Šemšić Sep 6 '16 at 15:24
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    $\begingroup$ I think you did that wrong. $$\dfrac{1*10^5}{22.414} = 4461$$ $\endgroup$ – MaxW Sep 6 '16 at 15:43
  • $\begingroup$ I used $T=298K$ and $p=101 325 Pa$ $\endgroup$ – Emir Šemšić Sep 6 '16 at 17:58
  • $\begingroup$ But still, we are off by a factor of 10. $\endgroup$ – Emir Šemšić Sep 6 '16 at 17:59
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Let's reformulate your equations a little bit. The ideal gas law was just fine: $$n = \frac{p V}{RT}$$ A problem lies in your formulation of Faradays law. We are not interested in masses, but in moles. Of course it is interchangeable by using the molar mass, but why make things complicated. As a rule of thumb you can remember for such tasks: Convert from mass in the beginning and convert to mass in the end (if needed). Nearly all equations in "the middle" are based on the concept of fixed relationsships between the number of involved particles and all equations are much easier to understand if you use $n$.

Faradays law means that the overall charge is just the product of charge per particle times the number of particles: $$ Q = nzF $$

The energy is just the product of $Q$ and $U$ divided by $\mu$. You don't need time and amperage. This gives: $$ E = \frac{QU}{\mu} = \frac{zFpVU}{RT\mu} $$

Unfortunately when putting values into these equations you get another result. The following python code will give 448.4 kWh as result. I think that your result is correct and our difference results from differently used constants...:

constants = {}
constants['F'] = 96485.33289
constants['R'] = 8.314
conversion = {}
conversion['to kWh'] = 2.7777e-07

n = lambda p, V, T : p * V / (constants['R'] * T)
Q = lambda n, z : n * z * constants['F']
E = lambda Q, U, mu : Q * U / mu
# Using p=1.01325*10^5 Pa, V=100m^3, T=298K, z=1, U=3.6V, mu=0.88
print(E(Q(n(1.01325e5, 100, 298), 1), 3.6, 0.88) * conversion['to kWh'], 'kWh')
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  • $\begingroup$ So, what result do you get when you run the above code? $\endgroup$ – Emir Šemšić Sep 15 '16 at 8:36
  • $\begingroup$ 48.9 kWh. Your desired result $\endgroup$ – mcocdawc Sep 15 '16 at 9:30
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    $\begingroup$ I get a different result see here: tinypic.com/r/x1isgm/9 $\endgroup$ – Emir Šemšić Sep 15 '16 at 17:50
  • $\begingroup$ I've also run the code - the answer generated by the code is 448.39. While the answer is wrong, I am not voting to delete this answer (who nominated it for deletion, btw???) because it needs to be debugged and is otherwise not "low quality." $\endgroup$ – Todd Minehardt Sep 15 '16 at 20:04
  • $\begingroup$ Perhaps it's all just a typo in the textbook, and that really is the answer. I hope we can come to a conclusion before the bounty runs out. $\endgroup$ – Emir Šemšić Sep 15 '16 at 20:41

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