(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

All textbooks

Brown 14th Edition

Ch.19 - Chemical Thermodynamics

Problem 97c

Chapter 19, Problem 97c

(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)

Verified step by step guidance

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.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

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.

Recommended video:

Guided course

01:14

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)

views

Textbook Question

The oxidation of glucose (C₆H₁₂O₆) in body tissue produces CO₂ and H₂O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C₂H₅OH) and CO2.

(a) Using data given in Appendix C, compare the equilibrium constants for the following reactions:

C₆H₁₂O₆(s) + 6 O₂(g) ⇌ 6 CO₂(g) + 6 H₂O(l)

C₆H₁₂O₆(s) ⇌ 2 C₂H₅OH(l) + 2 CO₂(g)

Textbook Question

The oxidation of glucose (C₆H₁₂O₆) in body tissue produces CO₂ and H₂O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C₂H₅OH) and CO2.

(b) Compare the maximum work that can be obtained from these processes under standard conditions.

C₆H₁₂O₆(s) + 6 O₂(g) ⇌ 6 CO₂(g) + 6 H₂O(l)

C₆H₁₂O₆(s) ⇌ 2 C₂H₅OH(l) + 2 CO₂(g)

(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)

Verified video answer for a similar problem:

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)