(b) Based on your general chemical knowledge, predict which of these reactions will have K>1. (i) 2 Mg(s) + O 2 (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I 2 (g) (iii) Na 2 (g) ⇌ 2 Na(g) (iv) 2 V 2 O 5 (s) ⇌ 4 V(s) + 5 O 2 (g)

1, the reaction favors the formation of products at equilibrium.>. 1 as they favor product formation.>. . .

(b) Based on your general chemical knowledge, predict which of these reactions will have K>1. (i) 2 Mg(s) + O 2 (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I 2 (g) (iii) Na 2 (g) ⇌ 2 Na(g) (iv) 2 V 2 O 5 (s) ⇌ 4 V(s) + 5 O 2 (g)

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.

Ch.19 - Chemical Thermodynamics

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Brown 15th Edition

Ch.19 - Chemical Thermodynamics

Problem 96b

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Chapter 19, Problem 96b

(b) Based on your general chemical knowledge, predict which of these reactions will have K>1. (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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<Understand the concept of equilibrium constant (K): The equilibrium constant, K, indicates the extent to which a reaction proceeds to form products. If K > 1, the reaction favors the formation of products at equilibrium.>

<Analyze reaction (i): 2 Mg(s) + O_2(g) ⇌ 2 MgO(s). This is a combustion reaction where magnesium reacts with oxygen to form magnesium oxide, a stable compound. Such reactions typically have K > 1 as they favor product formation.>

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (K)

The equilibrium constant (K) quantifies the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. A K value greater than 1 indicates that products are favored at equilibrium, while a K value less than 1 suggests that reactants are favored. Understanding K helps predict the direction of a reaction and its extent.

Gibbs Free Energy (ΔG)

Gibbs free energy (ΔG) is a thermodynamic potential that predicts the spontaneity of a reaction. A negative ΔG indicates that a reaction is spontaneous and will proceed in the forward direction, often correlating with a K value greater than 1. Conversely, a positive ΔG suggests non-spontaneity, aligning with a K value less than 1.

Phase States and Their Influence on K

The phase states of reactants and products (solid, liquid, gas) significantly influence the equilibrium constant. Only gases and solutes in solution are included in the K expression, while solids and liquids are omitted. Recognizing the phase states helps in determining the correct K value and understanding how changes in conditions affect the equilibrium position.

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Consider the following three reactions: (i) Ti(s) + 2 Cl₂(g) → TiCl₄(1g) (ii) C₂H₆(g) + 7 Cl₂(g) → 2 CCl₄(g) + 6 HCl(g) (iii) BaO(s) + CO₂(g) → BaCO₃(s) (c) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature.

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Using the data in Appendix C and given the pressures listed, calculate K_p and ΔG for each of the following reactions: (c) N₂H₄(g) → N₂(g) + 2 H₂(g) P_N2H4 = 0.5 atm, P_N2 = 1.5 atm, P_H2 = 2.5 atm

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Textbook Question

(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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Textbook Question

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane (C₂H₆), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: 2 CH₄(g) → C₂H₆(g) + H₂(g) In practice, this reaction is carried out in the presence of oxygen: 2 CH₄(g) + 12 O₂(g) → C₂H₆(g) + H₂O(g) (b) Is the difference in ΔG° for the two reactions due primarily to the enthalpy term (ΔH) or the entropy term (-TΔS)?

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The potassium-ion concentration in blood plasma is about 5.0⨉10^-3 M, whereas the concentration in muscle-cell fluid is much greater (0.15 M ). The plasma and intracellular fluid are separated by the cell membrane, which we assume is permeable only to K⁺. (a) What is ΔG for the transfer of 1 mol of K⁺ from blood plasma to the cellular fluid at body temperature 37 °C? (b) What is the minimum amount of work that must be used to transfer this K⁺?

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(b) Based on your general chemical knowledge, predict which of these reactions will have K>1. (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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Key Concepts

Equilibrium Constant (K)

Gibbs Free Energy (ΔG)

Phase States and Their Influence on K

(b) Based on your general chemical knowledge, predict which of these reactions will have K>1. (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)