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Homework answers / question archive / Problem 5
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Problem 5.74 Using values from Appendix C, calculate the value of ?H? for each of the following reactions. |
Part A CaO(s)+2HCl(g)?CaCl2(s)+H2O(g) Express your answer using four significant figures.
SubmitMy AnswersGive Up Part B 4FeO(s)+O2(g)?2Fe2O3(s) Express your answer using four significant figures.
SubmitMy AnswersGive Up Part C 2CuO(s)+NO(g)?Cu2O(s)+NO2(g) Express your answer using three significant figures.
SubmitMy AnswersGive Up Part D 4NH3(g)+O2(g)?2N2H4(g)+2H2O(l) Express your answer using five significant figures.
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To solve these problems, we use the following principle to obtain change in enthalpy of the reaction from standard enthalpies of formation from individual reactants and products:
*n is moles of the given reactant or product
For part A:
deltaHoH2O(g) ?= -241.8 kJ/mol deltaHoHCl(g) ?= -92.3 kJ/mol
deltaHoCaCl2(s) ?= -795 kJ/mol deltaHoCaO(s) ?= -635.5 kJ/mol
deltaH = [(-795 kJ/mol * 1 mol) + (-241.8 kJ/mol * 1 mol)] - [(-635.5 kJ/mol * 1 mol) + (-92.3 kJ/mol * 2 mol)]
deltaH =? -1080.8 kJ - (-820.1 kJ) = -216.7 kJ
For part B:
deltaHoFe2O3(s) ?= -822.2 kJ/mol
deltaHoFeO(s) ?= -272.2 kJ/mol
deltaH = [(-822.2 kJ/mol * 2 mol)] - [(-272.2 kJ/mol * 4 mol)]
deltaH =? -1644.4 kJ - (-1088.8 kJ) = -555.6 kJ
For part C:
deltaHoCu2O(s) ?= -166.7 kJ/mol deltaHoCuO(s) ?= -155.2 kJ/mol
deltaHoNO2(g) ?= 33.9 kJ/mol deltaHoNO(g) ?= 90.4 kJ/mol
deltaH = [(-166.7 kJ/mol * 1 mol) + (33.9 kJ/mol * 1 mol)] - [(90.4 kJ/mol * 1 mol) + (-155.2 kJ/mol * 2 mol)]
deltaH =? -132.8 kJ - (-220 kJ) = 87.2 kJ
For part D:
deltaHoH2O(l) ?= -285.8 kJ/mol deltaHoNH3(g) ?= -46.2 kJ/mol
deltaHoN2H4(g) ?= 95.4 kJ/mol
deltaH = [(-285.8 kJ/mol * 2 mol) + (95.4 kJ/mol * 2 mol)] - [(-46.2 kJ/mol * 4 mol)]
deltaH =? -380.8 kJ - (-184.8 kJ) = -196.00 kJ
