(c) In each case, indicate whether K should increase or decrease with increasing temperature. (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)

Hi everyone for this problem. It reads if the temperature of the dissociation of tetra phosphorous to die, phosphorus is increased by 120 Kelvin, will the equilibrium constant become larger or smaller and were given the reaction? Okay, so our question that we want to answer here is about the equilibrium constant. Okay. And whether or not it will become larger or smaller. Okay. So the first thing that we want to think about here is how does our equilibrium constant change with temperature. So our reaction can either be endo thermic or eggs. A thermic. Okay. And if it's endo thermic, that means we're going to treat heat as a reactant. Now, if our reaction is eggs are thermic, that means we're going to treat heat as a product. Okay, so in the problem we're told we're dealing with dissociation. Okay. And dissociation involves bond breaking, which is endo thermic. Okay, so we know that we have endo thermic reaction and that we're going to treat heat as a reactant. So let's go ahead and rewrite our reaction. So we have tetra phosphorous plus heat. Okay, so we're adding heat as a reactant, yields die phosphorus. Okay, now looking at this, we're told that if the temperature is increased. So increasing the temperature means we're adding heat and what outliers principle tells us is when stress is put on a system at equilibrium, the system is going to try to offset that stress. So by adding more heat, our equilibrium is going to want to shift away from that heat that was just added to offset that stress and so when it shifts away, that means we're going to be shifting to the right, Okay, so it's going to shift to the right to reach to offset that stress by shifting to the right, our concentration of reactant is going to decrease and our concentration of products is going to increase. So what does that mean in terms of our equilibrium constant? So our equilibrium constant, K is equal to our concentration of products over our concentration of react mints, products are in our numerator. Okay, so that means if we have a K that is large or greater than one, that means we have more product than reacting. So our K value increases as we get more product. On the other hand, if r K value is really small or less than one, that means we have more reactant over product now because we know we're shifting to the right, that means we're creating more product. And so that means that our equilibrium constant is going to become larger. Okay, so that is the answer to this problem. I hope this was helpful

Ch.19 - Chemical Thermodynamics

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Brown 15th Edition

Ch.19 - Chemical Thermodynamics

Problem 96c

Chapter 19, Problem 96c

(c) In each case, indicate whether K should increase or decrease with increasing temperature. (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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Identify the type of reaction for each case: exothermic or endothermic. This can often be inferred from the reaction equation or given data.

Apply Le Chatelier's Principle: For an exothermic reaction, increasing temperature shifts the equilibrium to the left, decreasing K. For an endothermic reaction, increasing temperature shifts the equilibrium to the right, increasing K.

For reaction (i) 2 Mg(s) + O2(g) ⇌ 2 MgO(s), determine if the formation of MgO is exothermic or endothermic. Typically, the formation of metal oxides is exothermic.

For reaction (ii) 2 KI(s) ⇌ 2 K(g) + I2(g), consider the phase change from solid to gas, which usually requires energy input, indicating an endothermic process.

For reactions (iii) Na2(g) ⇌ 2 Na(g) and (iv) 2 V2O5(s) ⇌ 4 V(s) + 5 O2(g), analyze the stoichiometry and phase changes to determine the enthalpy change, and apply Le Chatelier's Principle accordingly.

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Key Concepts

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

Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. This principle helps predict how the equilibrium constant (K) will respond to changes in temperature, pressure, or concentration. For example, if a reaction is exothermic, increasing the temperature will shift the equilibrium to favor the reactants, thus decreasing K.

Endothermic and Exothermic Reactions

Reactions can be classified as endothermic or exothermic based on their heat exchange with the surroundings. Endothermic reactions absorb heat, while exothermic reactions release heat. The temperature dependence of the equilibrium constant K is influenced by the nature of the reaction; for endothermic reactions, increasing temperature increases K, while for exothermic reactions, it decreases K.

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. It provides insight into the extent of a reaction and how it shifts in response to changes in conditions. Understanding how K varies with temperature is crucial for predicting the direction of a reaction's shift when temperature changes.

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Equilibrium Constant K

Related Practice

Textbook Question

Using the data in Appendix C and given the pressures listed, calculate K_p and ΔG for each of the following reactions:

(a) N₂(g) + 3 H₂(g) → 2 NH₃(g) P_N2 = 2.6 atm, P_H2 = 5.9 atm, P_NH3 = 1.2 atm

(b) 2 N₂H₄(g) + 2 NO₂(g) → 3 N₂(g) + 4 H₂O(g) P_N2H4 = P_NO2 = 5.0 × 10^-2 atm, P_N2 = 0.5 atm, P_H2O = 0.3 atm

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)

Textbook Question

(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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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)?

Textbook Question

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⁺?

Textbook Question

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)

(c) In each case, indicate whether K should increase or decrease with increasing temperature. (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

Le Chatelier's Principle

Endothermic and Exothermic Reactions

Equilibrium Constant (K)

(c) In each case, indicate whether K should increase or decrease with increasing temperature. (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)