Skip to main content
Ch.19 - Chemical Thermodynamics
Brown - Chemistry: The Central Science 14th Edition
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Brown 14th EditionCh.19 - Chemical Thermodynamics문제 97b
19장, 문제 97b

(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)

검증된 단계별 안내
1
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.
Analyze reaction (ii): 2 KI(s) ⇌ 2 K(g) + I_2(g). This reaction involves the decomposition of a solid into gaseous elements. Decomposition reactions of stable ionic solids like KI generally have K < 1, as they do not favor the formation of gaseous products.
Analyze reaction (iii): Na_2(g) ⇌ 2 Na(g). This reaction involves the dissociation of a diatomic molecule into atoms. The stability of Na_2 compared to individual Na atoms will determine K. Typically, such dissociation reactions have K < 1.
Analyze reaction (iv): 2 V_2O_5(s) ⇌ 4 V(s) + 5 O_2(g). This is a decomposition reaction of a stable oxide into its elements. Such reactions usually have K < 1, as they do not favor the formation of gaseous products from stable solids.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
4m
도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
가이드 코스
01:14
Equilibrium Constant K

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.
추천 영상:
가이드 코스
01:51
Gibbs Free Energy of Reactions

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.
추천 영상:
가이드 코스
03:19
States of Matter of Phase Diagrams
관련 실천
교과서 질문

(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)

교과서 질문

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

(a) N2(g) + 3 H2(g) → 2 NH3(g) PN2 = 2.6 atm, PH2 = 5.9 atm, PNH3 = 1.2 atm

(b) 2 N2H4(g) + 2 NO2(g) → 3 N2(g) + 4 H2O(g) PN2H4 = PNO2 = 5.0 × 10-2 atm, PN2 = 0.5 atm, PH2O = 0.3 atm

(c) N2H4(g) → N2(g) + 2 H2(g) PN2H4 = 0.5 atm, PN2 = 1.5 atm, PH2 = 2.5 atm

807
views
교과서 질문

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.

371
views
교과서 질문

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)

995
views
교과서 질문

(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)

860
views