Please help me with my he. I can't seem to solve this problem. I hope you can help me This is a guide to answer the activity. HESS LAW (Law of Heat Summation)“The enthalpy change of the overall process is the sum of the enthalpy changes of the individual component steps.” - It is the standard enthalpy of Reaction, ΔHᵒ. - It can be illustrated in the following example Consider the following reactions:(a) Fe(s) + O2(g) à FeO(s)ΔHᵒ = -272 kJ(b) 2 Fe(s) + O2 (g) à Fe2O3(s)ΔHᵒ = -82.5 kJ 1. Make sure to write the balance equation. 2. The final equation requires 2 moles of FeO which means we will multiply equation (a) and its enthalpy change by 2. 3. Reverse equation (a), which means, the product goes to the reactant side, (2) (FeO(s) à Fe(s) + O2(g))ΔHᵒ = (-272 kJ)(2) = 544 kJ2 FeO(s) à 2 Fe(s) + O2(g))ΔHᵒ = -825.5 kJ 4. Combine the final equation (a) and the initial equation (b). And then cancel the same components with same number of particles. In this case, cancel 2 Fe(s). (a) 2 FeO(s) à 2 Fe(s) + O2(g))ΔHᵒ = 544 kJ (b) 2 Fe(s) + O2 (g) à Fe2O3(s)ΔHᵒ = -825.5 kJ 5. Get the sum of the equation and its enthalpy. (a) 2 FeO(s) à 2 Fe(s) + O2(g))ΔHᵒ = 544 kJ (b) 2 Fe(s) + O2 (g) à Fe2O3(s)ΔHᵒ = -825.5 kJ ------------------------------------------------------------------------------- 2 FeO(s) + O2(g)) à Fe2O3(s)ΔHᵒ = 281.5 kJ Activity: Show complete solution Calculate the standard enthalpies of combustion at 25ᵒC of the following: (a) C4H8(g) + H2(g) à C4H10(g) ΔHᵒ = -2878.6kJ (b) C4H6(g) + 2H2(g) à C4H10(g) ΔHᵒ = -2543.5 kJ

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Please help me with my he. I can't seem to solve this problem. I hope you can help me This is a guide to answer the activity. HESS LAW (Law of Heat Summation) “The enthalpy change of the overall process is the sum of the enthalpy changes of the individual component steps.” - It is the standard enthalpy of Reaction, ΔHᵒ. - It can be illustrated in the following example Consider the following reactions: (a) Fe(s) + O2(g) à FeO(s)ΔHᵒ = -272 kJ (b) 2 Fe(s) + O2 (g) à Fe2O3(s)ΔHᵒ = -82.5 kJ 1. Make sure to write the balance equation. 2. The final equation requires 2 moles of FeO which means we will multiply equation (a) and its enthalpy change by 2. 3. Reverse equation (a), which means, the product goes to the reactant side, (2) (FeO(s) à Fe(s) + O2(g))ΔHᵒ = (-272 kJ)(2) = 544 kJ 2 FeO(s) à 2 Fe(s) + O2(g))ΔHᵒ = -825.5 kJ 4. Combine the final equation (a) and the initial equation (b). And then cancel the same components with same number of particles. In this case, cancel 2 Fe(s). (a) 2 FeO(s) à 2 Fe(s) + O2(g))ΔHᵒ = 544 kJ (b) 2 Fe(s) + O2 (g) à Fe2O3(s)ΔHᵒ = -825.5 kJ 5. Get the sum of the equation and its enthalpy. (a) 2 FeO(s) à 2 Fe(s) + O2(g))ΔHᵒ = 544 kJ (b) 2 Fe(s) + O2 (g) à Fe2O3(s)ΔHᵒ = -825.5 kJ ------------------------------------------------------------------------------- 2 FeO(s) + O2(g)) à Fe2O3(s)ΔHᵒ = 281.5 kJ Activity: Show complete solution Calculate the standard enthalpies of combustion at 25ᵒC of the following: (a) C4H8(g) + H2(g) à C4H10(g) ΔHᵒ = -2878.6kJ (b) C4H6(g) + 2H2(g) à C4H10(g) ΔHᵒ = -2543.5 kJ

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