## 9.67

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

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

ClaireHW
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### 9.67

Assume that the standard change in enthalpy and entropy are independent of temperature and calculate the change in the standard Gibbs Free Energy for each of the following at 80 C. Over what temperature range will each reaction be spontaneous under standard conditions?
a) B2O3 + 6HF -> 2BF3 + 3H2O
b) CaC2 + 2GCl -> CaCl2 + C2H2
c) C(graphite) -> C(diamond)

(Claire Woolson Dis 1K)

Mika Sonnleitner 1A
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### Re: 9.67

For this problem, you need to find the temperature at which the reaction is spontaneous, so you first have to find ΔG. For part (a), you find ΔG by first finding ΔH and ΔS.

Step 1: Find ΔG° by finding ΔH° and ΔS°, and using the equation ΔG°= ΔH°-TΔS°.
You can look up the values in Appendix 2A, using the formula ΔH/ΔS= sum of the enthalpy/entropy of products-reactants. In part (a), these values are ΔH°=-232.1 kJ/mol, ΔS°=-378.68 J/Kmol. Plug them into the equation to find ΔG°=-98.37 kJ/mol

Step 2: To find the temperature at which the reaction is spontaneous, you need to find the temperature value that causes ΔG° to be negative. We do this by setting ΔG°=0. (Don't forget to convert ΔH° to J/mol or convert ΔS° to kJ/Kmol)
0=ΔH°-TΔS°
0=-23211 J/mol -T(-378.68 J/Kmol)
T=612.92 K

Step 3: Since both ΔH° and ΔS° are negative, the reaction is spontaneous at lower temperatures. So, when T<621.9 K.

repeat this process for parts (b) and (c).

zanekoch1A
Posts: 25
Joined: Thu Jul 13, 2017 3:00 am

### Re: 9.67

So you have to combine the deltaH of products and reacants and deltaS of products and reactants before you put them into the Gibbs equation?