(a) For each of the following reactions, predict the sign of ΔH° and ΔS° without doing any calculations. (i) 2 Mg(s) + O₂ (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I₂(g) (iii) Na₂(g) ⇌ 2 Na(g) (iv) 2 V₂O₅(s) ⇌ 4 V(s) + 5 O₂(g)

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Hello everyone. So in this video we're given four different statements and without doing any calculations, we're trying to find the signs of our delta H. So the change in entropy and the signs are delta S. The change of our entropy. So first we can talk a little bit about the meaning behind the different signs and then once we get to know that we can go ahead and apply it by looking at the equation itself. So we can use this as a kind of a key. So if we have a negative delta H, this means that we are forming bonds and if there's a face change, we're going from gas to liquid to solid in this order and not the other way around. But if we're dealing with a positive delta H, this means that bonds are breaking and that the phase change can be from solid to liquid. And then to gas. So basically the reverse if of if we have a negative delta H now for our entropy, our delta us if that is Greater than zero, basically a positive delta us. This means that our entropy is increasing Now if our delta s is less than zero, basically we have a negative delta S than our entropy is decreasing. Now playing this key here to our four different statements, let's go ahead to go look. So we look at the first one we see H2 going to two moles of hydrogen. We can see that bonds are breaking here. So that means we have a positive delta H. As for delta us, we see that we're going from one mole of gas to two moles of gas. So we're increasing the entropy. So we have our positive delta S. Now for our 2nd we have also again bonds breaking. We have one product here but we have two or one starting region here, but we have two products. So bonds are breaking again, then we have a positive let's write this baby on top, We have a positive delta H. And then for our delta S. Let's see here. So we are starting off with two moles of solid and then we're going to four moles of solid and then three moles of gas. So we're going from two moles basically to seven rolls are increasing the entropy. So we have a positive delta s. Now for statement number three we see that bonds are forming because we're starting out with two different molecules and then we have one product. So bonds are forming, therefore we have a negative delta H. And then for our entropy, let's see here we're starting off with two moles of solid and one mole of gas. And then we're ending with two moles of solid. So we're going from three moles are the starting materials down to two moles. So it means we're decreasing entropy. So we have a negative delta s. Now lastly our for statement let's see here. So we have one product, then we associate into two different products. So our bonds are breaking here, that means that we have a positive delta H. And as for change in entropy, let's see here we're starting off two moles of solid. Then we have two more solid on the product side, as well as one mole of gas. So we're going from two most to three moles. That means we're increasing entropy and therefore we have a positive delta S. Alright, so we have our signs of our different reactions here without doing any calculations. And those are final answers for this problem.

(a) For each of the following reactions, predict the sign of ΔH° and ΔS° without doing any calculations. (i) 2 Mg(s) + O₂ (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I₂(g) (iii) Na₂(g) ⇌ 2 Na(g) (iv) 2 V₂O₅(s) ⇌ 4 V(s) + 5 O₂(g)

Verified Solution

Key Concepts

Enthalpy (ΔH)

Entropy (ΔS)

Phase Changes and Reaction Direction

(a) For each of the following reactions, predict the sign of ΔH° and ΔS° without doing any calculations. (i) 2 Mg(s) + O2 (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I2(g) (iii) Na2(g) ⇌ 2 Na(g) (iv) 2 V2O5(s) ⇌ 4 V(s) + 5 O2(g)

Verified Solution

Key Concepts

Enthalpy (ΔH)

Entropy (ΔS)

Phase Changes and Reaction Direction

(a) For each of the following reactions, predict the sign of ΔH° and ΔS° without doing any calculations. (i) 2 Mg(s) + O₂ (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I₂(g) (iii) Na₂(g) ⇌ 2 Na(g) (iv) 2 V₂O₅(s) ⇌ 4 V(s) + 5 O₂(g)