At what temperatures is the following reaction, the reduction of magnetite by graphite to elemental iron, spontaneous? Fe₃O₄(s) + 2 C(s, graphite) → 2 CO₂(g) + 3 Fe(s)

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Hello everyone. So in this video we're trying to determine the temperature at which the reaction of Lead oxide being reduced by carbon or graphite to produce light metal occurs. So let's go ahead and figure this out. First thing is to assume that we are in standard temperature conditions, the center temperature in Kelvin's would be 298 Calvin's. Let's go ahead and write this out. So we're assuming that the temperature is equal to 2 98 kelvin's. Okay. First things first is go ahead and calculate for our delta H. So our change in entropy. So the delta H is equal to the delta H of our products minus the delta age of our reactant. So delta H. And I'm just using values from this given table right over here and we're concerning this chemical equation right over here. Alright, so we see on the product side we have the two moles of PB and one mole over C. 02. Right, so we again have the two moles, the corresponding delta H. For that is going to be zero killer jewels per mole. And then my second product, we have one mole of C. 02 corresponding delta H is going to be negative 93.5 units being killed, joules per mole. And then to do for delta age for my reactant. So we have again, just two moles of our P B O. And then one more of our carbon. Okay, so corresponding delta H value is negative 2 17. killer jewels per mole. And then my other reactant is going to be. Again we have one mole of that and then the corresponding delta H value is zero killer jewels per mole. Now putting all these values into my calculator, I get that the delta H. Is going to be 41.1 killer jewels. So we have a delta H. Let's go ahead and solve our delta S. The change of entropy. So my delta S is going to be the entropy of my reactant minus the entropy of the products. This kind of flipped in the sense of comparing it to our delta H. So same process though. So Delta H. And delta S. Again we have two moles Multiplied by 68.85 jewels per mole times Kelvin plus one more corresponding entropy value, we have to 13.6 joules per mole times kelvin, So that's going to be of my reactant. Now we do the entropy of the products, so we again have our two moles multiplied with 68.70 joules per mole times kelvin. And then we're gonna add AC. We have one mole and then the corresponding entropy value is 5.69 units being jewels per mole times Kelvin. So slightly off centered here but we know that value already, so we just ended up there. Alright I'm putting all these values into my calculator, I get them, I delta S is equal to 208.21 jules per kelvin. We actually want the units being in kilo jewels instead of our jewels. So just do a direct conversion from jules and two kg jewels. So for everyone, kill a jewel that we have, we have 1000 jewels. Again, crossing out that jules. Alright, so our delta S. Then the change of entropy is going to be equal to 0.28 to one kg joules per kelvin. Again, just scrolling down to give us more space. Here, we're gonna go ahead and solve our delta gino. The delta G is equal to the delta G of our products. When I see delta G of my reactant again, just using the table above. That's given to us from the problem. So we have two moles of this first product corresponding delta G value is zero kila jewels per mole they're adding with the second product. So one mole that corresponding delta G is negative 394.4 kg joules per mole. And then now for the delta of the reactant. So we have two moles and then negative 187.9. Kill a jules Permal. We're adding the delta G. Of our second reactant. We have one mole of that corresponding delta G is zero killer jewels per mole. So putting all that into my calculator, I get that my delta G value is negative 18.6 kg jewels. And of course because of that value, we can see that the sign of delta G is negative. So it's going to be spontaneous at the center condition temperature of 298 kelvin's let's go ahead now calculate for our tea range, our temperature range, so T. Range. So of course we have the gibbs free energy equation being delta G. Equaling to the delta H minus T. Multiplied by the change of entropy which is delta us are just A G. Will be equal to zero. We're solving for T. Here we get that T. Is equal to well delta H over delta S plug in those values that are known and solved already. We have 41.1 killer jewels and then delta S. Is 0.2821 killer jewels per kelvin. And we can see of course that the killer jewels units will cancel giving us a T. Value Of 7.4 Kelvin's. And we want this in Celsius. So we'll subtract To 73.15 to give us that. The T. is equal to negative 75.7°C. So the reaction is spontaneous at temperatures greater than our negative 75.7°C. Let's go ahead and write this out. So our final answer again is that the reaction is spontaneous At temperatures greater than negative 75.7°C. So this is going to be my final answer for this problem. Thank you all so much for watching

At what temperatures is the following reaction, the reduction of magnetite by graphite to elemental iron, spontaneous? Fe₃O₄(s) + 2 C(s, graphite) → 2 CO₂(g) + 3 Fe(s)

Verified Solution

Key Concepts

Gibbs Free Energy

Entropy and Enthalpy

Temperature's Role in Spontaneity

At what temperatures is the following reaction, the reduction of magnetite by graphite to elemental iron, spontaneous? Fe3O4(s) + 2 C(s, graphite) → 2 CO2(g) + 3 Fe(s)

Verified Solution

Key Concepts

Gibbs Free Energy

Entropy and Enthalpy

Temperature's Role in Spontaneity

At what temperatures is the following reaction, the reduction of magnetite by graphite to elemental iron, spontaneous? Fe₃O₄(s) + 2 C(s, graphite) → 2 CO₂(g) + 3 Fe(s)