Problem

Using the relevant $S_{298}^{\circ}$ values liste…

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Problem 24 Hard Difficulty

"Thermite "reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Fe}(s)$ . Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 $\mathrm{kJ} / \mathrm{mol}$ of heat.

Answer

The reaction is spontaneous.

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Video Transcript

the question asked us to find out whether the given in the action is spontaneous or not, and it starts off by telling us the change in entropy of the reaction which is negative. 851.8 tillage joules per mole, which we can convert into negative 851,800 jewels for more in order to find if the reaction is spontaneous or not, we actually need to determine the value of Delta G, where the changing gives free energy, understanding conditions off the reaction and that we can do this association Since we already know the change in entropy adult age standard. We know the temperature which on this reaction operates under its standard temperature of 298 Kelvin and we're meant to now find Delta s or the change in entropy, understand or conditions of the reaction. We can find this change in entropy of the reaction understanding conditions by doing the by finding the difference off between the products and reactant off the standard n trapeze. And so here we're gonna basically work based off this formula, and we're going to get the standard entropy values from appendix Gene in the textbook. And once we get those values were going to end up with, um sorry. Okay. Worker and up with 50 0.92 plus two times 27.3 were multiplying the stander entropy off iron by two. Because in the balance to reaction, given the question has a coefficient of two. So remember to take these coefficients into account. Um, Now we're gonna do minus 87.4 plus two times 28.3. Once we do all the math rules, we should get a value of negative 38 0.48 as the the standard changed entropy for the reaction. With this, we can go ahead and move on to the next step of finding Delta G standard off the reaction and so that we can do by the change in and throw P, which is negative. 851 800 minus. The temperature was just to 98 kelvin times. Uh, the changing entropy, which is negative. 38.48 This is the value that we just found in the previous stuff. Um, and this is going to equal. Okay. Changing gives free energy understanding conditions for the reaction. Ah, once this is calculated, we find out that this is going to be negative. 86 three, 267 Jewels. And because our answer is negative, that means that Delta G standard for the reaction is going to be less than zero, and that makes the reaction spontaneous.

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