Question

0 In K vs 1/T With Trendline ol4 V be y6866.30x-9.3041 R2-0.9992 30.00 -30,50 31.00 31.50 32.00 32.50 33.00 -33.50 -34.00 -34.50 -35.00 3.00 3.70 3.60 3.50 3.40 3.30 3.20 3.10 1/T (x 10 K) 4. Recalling that A, G = -RTlnKeq , use the plot shown to find the Enthalpy change for the auto-ionization of water at the temperature of the human body (37°C). Find the Change in entropy for the self-ionization of water at this temperature. to notuloz Iclom 00 to omulov 5. The boiling temperature of cyclohexane, CsH12, is 354 K. Use Trouton's rule to estimate the enthalpy of vaporization of cyclohexane. 6. Use the following data to calculate the standard reaction entropy at 298 K for the thermal decomposition of ammonium chloride, NH4CI. Give two reasons for why the change in entropy is so large and in the direction it is. sheids no vio9ibhow aniog gniflom NH&CI(s)NH3(g) + HCI(g) NH4CI(s) 94.85 NH3(g) HCI(g) Sm /Jmol1 192.77 186.90 In K

Answer 1

4. The Van't Hoff equation states that:

${\displaystyle {\frac {d\ln K_{\mathrm {eq} }}{d{\frac {1}{T}}}}=-{\frac {\Delta H^{\ominus }}{R}}.}$

Which means that, in a graph of ln(K) vs 1/T, the slope has a value equal to -∆H/R. In this case, we have:

$-6866.30K=-\frac{\Delta H}{R}$

So:

$\Delta H=6866.30K\cdot 8.314\frac{J}{mol\cdot K}=57086\frac{J}{mol}$

From the linear fit we can calculate K at 37° (310 K):

$ln(K)=-6866.30K\frac{1}{T}-9.3041=-6866.30K\frac{1}{310K}-9.3041=-31.45$

$K=e^{-31.45}=2.19x10^{-14}$

With this, we can calculate the standard free gibbs energy:

$\Delta G^{o}=-RTln(K)=-8.314\frac{J}{molK}\cdot 310K\cdot ln(2.19x10^{-14})=81066J/mol$

And we can now calculate the change in entropy using:

$\Delta G^{o}=\Delta H^{o}-T\Delta S^{o}$

So:

$\Delta S^{o}=-\frac{\Delta G^{o}-\Delta H^{o}}{T}=-\frac{81066\frac{J}{mol}-57086\frac{J}{mol}}{310K}=77.4\frac{J}{mol\cdot K}$

5. Trouton rule states that the entropy of vaporization of a liquid can be estimated as 10.5R. So, at the boiling point, where liquid and gas are in equilibrium (which means that delta G = 0), we have:

$\Delta G=0=\Delta H-T\Delta S$

So we can estimate the enthalpy of vaporization as:

$\Delta H=T\Delta S=354K\cdot 10.5\cdot 8.314\frac{J}{mol\cdot K}=30903\frac{J}{mol}$

6. The standard reaction entropy of a reaction can be calculated as the difference between the standard entropy of the products and of the reactants. In this case:

$\Delta S^{o}=S_{NH_{3}}^{o}+S_{HCl}^{o}-S_{NH_{4}Cl}^{o}=(192.77-186.90-94.8)\frac{J}{mol\cdot K}=284.87\frac{J}{mol\cdot K}$

It is positive, so it is "in the direction" of decomposition. It has such a large value because, if we consider entropy in its most primitive definition of "disorder", we are decomposing a solid (a very ordered state of matter) into gases (a state that has very little order). So, overall, we are increasing the disorder of the system.