hw1solution_pdf - casey(rmc2555 Homework 1 holcombe(51395...

This preview shows page 1 - 3 out of 9 pages.

casey (rmc2555) – Homework 1 – holcombe – (51395)1Thisprint-outshouldhave26questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.0013.8pointsThe direct vaporization of a solid withoutpassing through its liquid state is called7.66 kJ/molExplanation:ΔH= 81 kJBonds broken:3 NBr bonds atxkJ each6 OH bonds at 459 kJ eachBonds made:3 NH bonds at 386 kJ each3 OH bonds at 459 kJ each3 OBr bonds at 201 kJ each.1.sublimation.correct2.condensation.3.solidification.4.evaporation.Explanation:Sublimation is the direct vaporization of asolid without passing through its liquid state.This is a physical change.Solidsublimation----------→gasSublimation is an endothermic process.0023.8pointsThe energy change ΔHassociated with thereactionNBr3(g) + 3 H2O(g)3 HOBr(g) + NH3(g)is +81 kJ/mol rxn.These bond energy values might be useful:O-H 459 kJ/mol; N-H 386 kJ/mol; O-Br 201kJ/mol.The strength of the NBr bond is1.128 kJ/mol2.4 kJ/mol3.248 kJ/mol4.155 kJ/molcorrect5.280 kJ/mol6.465 kJ/molΔH= (bonds broken)-(bonds made)81 = [3x+ 6(459)]-[3(386) + 3(459) + 3(201)]81 = 3x+ 2754-31383x= 465x= 155 kJ0033.8pointsCalculate the standard reaction enthalpy forthe reactionNO2(g)NO(g) + O(g)Given:O2(g)2 O(g)ΔH= +498.4 kJ/molNO(g) + O3(g)NO2(g) + O2(g)ΔH=-200 kJ/mol3O2(g)2 O3(g)ΔH= +285.4 kJ/mol.1.+355 kJ/mol2.+413 kJ/mol3.+592 kJ/mol4.+192 kJ/mol5.+555 kJ/mol6.+306 kJ/molcorrect7.+93.5 kJ/molExplanation:
casey (rmc2555) – Homework 1 – holcombe – (51395)2O2(g)2 O(g)will be exothermic or endothermic?the speed of the reaction2.the heat properties of the products3.the activation energy of the reaction4.the relative concentrations of reactantsand products in solution5.the difference in energy between reactantsand productscorrectExplanation:ΔHis the energy difference between theproducts and reactants; ΔHfor exothermicreactions is negative, and ΔHfor endother-mic reactions is positive.0063.8pointsThe standard heat of formation for sulfurdioxide gas is 296.8 kJ/mol.Calculate theamount of energy given off when 31 g ofSO2(g) is formed from its elements.Correct answer: 143.538 kJ.Explanation:ΔH0f SO2=-296.8 kJ/molm = 31 gΔH0ffor 31 g SO2Which of the following statements is trueabout an exothermic reaction?1.We can determine the absolute heat con-tent of the products and reactants.3.The change in enthalpy is positive.0053.8pointsWhat factor determines whether a reactionwill be exothermic or endothermic?1.the speed of the reaction2.the heat properties of the products3.the activation energy of the reaction4.the relative concentrations of reactantsand products in solution5.the difference in energy between reactantsand productscorrectExplanation:ΔHis the energy difference between theproducts and reactants; ΔHfor exothermicreactions is negative, and ΔHfor endother-mic reactions is positive.0063.8pointsThe standard heat of formation for sulfurdioxide gas is 296.8 kJ/mol.Calculate theamount of energy given off when 31 g ofSO2(g) is formed from its elements.Correct answer: 143.538 kJ.Explanation:ΔH0f SO2=-296.8 kJ/molm = 31 gΔH0ffor 31 g SO2

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture