From the values given for ΔH° and ΔS°, calculate ΔG° for each of ...

Question

From the values given for ΔH° and ΔS°, calculate ΔG° for each of the following reactions at 298 K. If the reaction is not spontaneous under standard conditions at 298 K, at what temperature (if any) would the reaction become spontaneous?
(a) 2 PbS1s2 + 3 O21g2¡2 PbO1s2 + 2 SO21g2 ΔH ° = -844 kJ; ΔS° = -165 J>K

Accepted Answer

Hello everyone. So in this video we have two parts. The first part is to calculate for the delta G. Of the reaction that's given to us by this chemical reaction over here. The second part is once we calculate delta G. Is the reaction going to be non spontaneous or not? If it is not spontaneous. And we have to go ahead and identify the temperature if any. I wish the reaction is spontaneous. So let's go ahead and bring out gibbs free equation or free energy equation. As recall, that's going to be delta G. Equaling to the delta H. One is the T. For temperature. Multiply with our delta S. So we have all the value is given to us in the problem. So delta G is equaling two negative 900.0 killer jewels minus R. T. Value is 2 98 kelvin's. Delta S is let's see, we have negative 151.5 joules per kelvin. Let's actually convert that jules into killer jewels with a direct conversion. So for every one killer jewels we have 1000 jewels putting into the calculator. My numerical values here for this equation, we get the value of negative 855 kg jewels. And we see here that this reaction is going to be spontaneous at standard conditions because we have this calculated value. So here we again, it's going to be spontaneous at standard conditions. So this over here is going to be my final answer for this problem. Thank you all so much for watching

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