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
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
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Chapter6: Thermochemisty
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Problem 6.147QP: The head of a strike anywhere match contains tetraphosphorus trisulfide, P4S3. In an experiment, a...
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The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
Question
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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