## practice test 6A&6B

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

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

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

Ya Gao
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am
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### practice test 6A&6B

The question states that 'We expect a reaction to be spontaneous if it is exothermic at a very low temperature.' I wonder if we need to know the sign of ∆S to figure out if the ∆G is negative.For the next true or false question, it said that We can expect an endothermic reaction to be favorable at a vert high temperature. I don't get it why it is false. Some clarification would be helpful.

Samantha Kan 2L
Posts: 30
Joined: Sat Jul 22, 2017 3:00 am

### Re: practice test 6A&6B

For 6a), the reaction is exothermic, so delta H is negative. We are also given that the temperature is very low. Based on the equation delta G = delta H - T(delta S), if delta S is positive, then the reaction is spontaneous since the answer will be negative. Even if delta S is negative, since the temperature is very low, the answer will still be negative, which is why that statement was true.

For 6b), the reaction is endothermic, so delta H is positive. The temperature is very high. Based on the equation delta G = delta H - T(delta S), if delta S is positive, then the reaction would be spontaneous. However, if delta S is negative, the reaction would not be spontaneous, since the answer would be positive. Therefore, for this scenario, we need to know the sign of delta S.

Hope this helps!