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I have the reaction $\ce{2COCl2(g) -> CCl4(l) + CO2(g)}$

And I want to calculate $T$ (temperature) when $Q = 10^{-4}$ at equilibrium.

I tried simply putting $$T = -\frac{\Delta G^{0}}{\Delta S^{0}-R \ln Q}$$

Anyway, since $\Delta G^{0}$ for the reaction ($\pu{42000J/mol}$) is the value at standard condition (25 degrees celsius) I use that $\Delta G = \Delta H^{0} - T\times\Delta S^{0} + RT\ln Q$ where $\Delta G = 0$ since we're at equilibrium.

From this I obtain that $$T = \frac{\Delta H^{0}}{\Delta S^{0}-R\ln Q}$$

But which $\Delta H^{0}$ and $\Delta S^{0}$ values do I use here? I can calculate them from the standard conditions at 25 degrees celsius using SI, but that wouldnt be correct, would it? I'm rather confused by this task, which seems simple at first sight. Hope someone can clarify.

(The answer should be $\pu{476K}$)

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  • $\begingroup$ Are you familiar with the Van't Hoff equation? $\endgroup$ Oct 31, 2017 at 11:37
  • $\begingroup$ No, not yet. Is this required here? $\endgroup$
    – novo
    Oct 31, 2017 at 11:39
  • $\begingroup$ Your $\LaTeX$ skills are fine, as far as $\Delta{}G^0$ goes. There was another thing that needed correction, but it is specific to this site. $\endgroup$ Oct 31, 2017 at 12:05
  • $\begingroup$ Look it up online and see. $\endgroup$ Oct 31, 2017 at 12:39
  • $\begingroup$ This question was taken from an earlier test I'm practising on, van't Hoff shouldn't be required. Is there a way around it without using van't Hoff? $\endgroup$
    – novo
    Oct 31, 2017 at 14:10

1 Answer 1

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We know that, at temperature T, $$\Delta G^0(T)=\Delta H^0(T)-T\Delta S^0(T)$$ From Hess's Law, we also know that $$\Delta H^0(T)=\Delta H^0(298)+???$$ and $$\Delta S^0(T)=\Delta S^0(298)+???$$

That should be a good enough hint.

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