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A maximum of $1.89~\mathrm{mg}$ of $\ce{AgCl}$ will dissolve in $1.00~\mathrm{L}$ of solution at $25.00~^\circ\mathrm{C}$. A maximum of $5.26~\mathrm{mg}$ of $\ce{AgCl}$ will dissolve in $1.00~\mathrm{L}$ of solution at $50.00~^\circ\mathrm{C}$. Use this information to calculate $\Delta H^\circ$ and $\Delta S^\circ$ for the following reaction. $$\ce{AgCl~(s) -> Ag+ (aq)~+ Cl- ~(aq)}$$ Assume that ΔH and ΔS are independent of temperature.

I started off by converting the given masses into moles using molar mass of $\ce{AgCl}$ and since we're given the reaction, we can assume that $K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]$, thus

  • At $25~^\circ\mathrm{C}$, after doing conversion I got $1.32\cdot 10^{-5}~\mathrm{mol}$ of $\ce{AgCl}$

  • At $50~^\circ\mathrm{C}$ I got $3.67\cdot 10^{-5}~\mathrm{mol}$ $\ce{AgCl}$

Since, $K_\mathrm{sp}=[\ce{Ag+}][\ce{Cl-}]$ and mole ratio is 1:1, I calculated the two $K_\mathrm{sp}$ values:

  • At $25~^\circ\mathrm{C}$: $K_\mathrm{sp1} = 1.74 \cdot 10^{-10}$
  • At $50~^\circ\mathrm{C}$: $K_\mathrm{sp2}= 1.35\cdot 10^{-9}$

Now, we do have two temperatures and two equilibrium constants which empowers me to use this formula: $$\ln\left(\frac{K_1}{K_2}\right)=-\frac{\Delta H}{R}\cdot \frac{T_2-T_1}{T_1T_2}$$

After solving for $H^\circ$ I get $65.6~\mathrm{kJ/mol}$ which is the correct answer.

How do I find $\Delta S$? I tried calculating it via standard $$\Delta G= \Delta H - T\Delta S$$ and averaging but it does not work.

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    $\begingroup$ actually I figured it out, false alarm . If anybody stumbles upon a problem like this you should use this formula : Ln(K)=-(dH/RT)+(dS/R) which gives the correct answer and ties in K, dH all together $\endgroup$ – Jx1 Jun 3 '15 at 2:32
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    $\begingroup$ Would you care to answer your own question? It is highly encouraged in the Stack Exchange network! $\endgroup$ – Nicolau Saker Neto Jun 3 '15 at 2:56

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