Hess's Law

Calculate the enthalpy of the following reaction:

Using Hess’s Law, determine the ΔH of the following reaction: N2(g) +  2O2(g)  →  2NO2(g) Given the following equations N2(g)  +  3H2(g)  →  2NH3(g)                                       ΔH = -115 kJ 2NH3(g)  +  4H2O(l)  →  2NO2(g)  +  7H2(g)                ΔH = -142.5 kJ H2O(l)  →  H2(g)  +  1/2O2(g)                                      ΔH = -43.7 kJ        

Researchers at Purdue University have exploited the reactivity of a reaction to develop a novel type of rocket fuel consisting of aluminum nanoparticles dispersed in H2O(s). They nicknamed this propellant “ALICE” (get it? Al-ice) and noted that in addition to being environmentally friendly, ALICE could be produced on any extraterrestrial body that contains water. The recent discovery of water on Mars could draw new attention to this intriguing fuel. (International Journal of Aerospace Engineering, 2012) One can speculate that the reaction that causes ALICE to act as a propellant is: 2Al (s) + 3H2O(s) → Al2O3(s) + 3H2(g)   where the hydrogen gas is subsequently burned.  A. Given the molar enthalpies for the following reactions,  2Al(s) + 3/2 O2 (g) → Al2O3 (s)        ΔH° = -1669.8 kJ/mol   H2 (g) + 1/2 O2 (g) →  H2O (l)         ΔH° =    -285.8 kJ/mol H2O (s) → H2O (l)                           ΔH° =       6.01 kJ/mol determine the ΔH° for the ALICE reaction 2Al (s) + 3H2O (s) → Al2O3 (s) + 3H2 (g)       Is the ΔH° you calculated consistent with Al-ice serving as a fuel? ________________ Give one reason to support your answer.    

Given the following reactions Fe2O3 (s) + 3CO (s) → 2Fe (s) + 3CO2 (g) ∆H = -28.0 kJ 3Fe (s) + 4CO2 (s) → 4CO (g) + Fe 3O4 (s) ∆H = +12.5 kJ the enthalpy of the reaction of Fe 2O3 with CO 3Fe2O3 (s) + CO (g) → CO2 (g) + 2Fe3O4 (s) is ________ kJ.   a) 40.5 b) +109 c) -15.5 d) -109 e) -59.0

Given the following reactions N2 (g) + 2O2 (g) → 2NO2 (g)   ΔH = 66.4 kJ 2NO (g) + O2 (g) → 2NO2 (g)  ΔH = -114.2 kJ the enthalpy of the reaction of the nitrogen to produce nitric oxide is ________ kJ.   a) -47.8 b) 47.8 c) 180.6 d) -180.6 e) 90.3

Calculate ΔH° (in kJ) for the reaction. 2S (s) + 3O2 (g) → 2SO3 (g)  ∆H = -790 kJ S (s) + O2 (g) → SO2 (g)        ∆H = -297 kJ the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide is ________ kJ.   a) -196 b) -543 c) 1087 d) 196 d) -1384

The United States gets 95% of its hydrogen gas by "steam reforming" methane, CH4(g). Another option is the “partial oxidation” of hydrocarbons to make carbon monoxide and hydrogen gas. For octane, that reaction would be  C8H18(l) + 4O2(g) → 8CO(g) + 9H2(g)   We know ΔH° for the combustion of octane  C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(l)      ΔH° = -5500 kJ/mol  Assume the arbitrary values below for the combustion reactions of carbon monoxide and hydrogen: CO(g) + 1/2 O2(g) → CO2(g)             ΔH° = “-290 kJ/mol”  H2(g) + 1 /2O2(g) → H2O(l)               ΔH° = “-285 kJ/mol” A. Do you expect the magnitude of ΔH° for the partial oxidation of octane, to be greater or less than the magnitude of ΔH° for the combustion of octane? Explain.     B. Now, determine   ΔH°  for the partial oxidation of octane with those assigned values. Choose from the choices below         C. Do you think the hydrogen produced provides more or less energy than you would get from the combustion of octane? Explain. 

Given these values of ΔH° : CS2 (l) + 3O2(g) → CO2(g) + 2SO2(g)            Δ H° = -1077 kJ H2 (g) + O2 → H2O2(l)                                    Δ   H° = -188 kJ                 H2 (g) 1/2 O2(g) → H2O(l)                              Δ   H° =  -286 kJ What is the value of  ΔH° for this reacion? CS2(l) + 6H2O2(l) → CO2(g) + 6H2O (l) + 2SO2(g) a) -1175 kJ•mol-1 b) -1151 kJ•mol-1 c) -1665 kJ•mol-1 d) -3921 kJ•mol-1

Given the chemical reactions: 2 NO2(g) → N2(g) + 2 O2(g)                  ΔH = −68 kJ N2O4(g) → N2(g) + 2 O2(g)                   ΔH = −10 kJ Determine the ΔH for the following reaction: N2O4(g) → 2 NO2(g)   A. −78 kJ B. −58 kJ C. 48 kJ D. 58 kJ E. 78 kJ

The values for the enthalpy of reaction are given for two reactions below Mg(s) + Na 2CO3(s) → 2 Na(s) + MgCO 3(s)           ΔH rxn = 317.1 kJ   Mg(s) + 2 NaF(s) → 2 Na(s) + MgF 2(s)                ΔH  rxn = 35.6 kJ   Based on this information, what is the value for ΔH rxn for the reaction MgCO3(s) + 2 NaF(s) → Na 2CO3(s) + MgF2(s) a. + 352.7 kJ b. + 281.5 kJ c. + 245.9 kJ d. - 281.5 kJ e. - 352.7 kJ

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: P4(g) + 10Cl2(g) → 4PCl5(s)                ΔH°rxn = ? Given PCl5(s) → PCl3(g) + Cl2(g)                   ΔH°rxn = +157kJ P4(g) + 6Cl2(g) → 4PCl3(g)                  ΔH°rxn = -1207kJ A) -1835 kJ B) - 1364 kJ C) -1050 kJ D) -1786kJ E) -2100 kJ

Given the following data: 3CO2(g) + 4H2O(l) ⇌ C3H8(g) + 5O2(g) ∆H = 1, 110 kJ · mol−1 H2O(l) ⇌ H2(g) + 1/2O2(g) ∆H = 142.5 kJ · mol−1 3Cgraphite + 4H2(g) ⇌ C3H8(g) ∆H = −52 kJ · mol−1 calculate ∆H for the reaction Cgraphite + O2(g) ⇌ CO2(g) 1. 1202 kJ · mol−1 2. −592 kJ · mol−1 3. 543 kJ · mol−1 4. −197 kJ · mol−1 5. −543 kJ · mol−1  

Consider the combustion of liquid methanol, CH 3OH (l):CH3OH (l) + 3/2 O2 (g) → CO2 (g) + 2 H2O (l)     ΔH = -726.5 kJ(a) What is the enthalpy change for the reverse reaction? Explain. 

Consider the combustion of liquid methanol, CH 3OH (l):CH3OH (l) + 3/2 O2 (g) → CO2 (g) + 2 H2O (l)     ΔH = -726.5 kJ(b) Balance the forward reaction with whole-number coefficients. What is ΔH for the reaction represented by this equation? Explain.

Consider the decomposition of liquid benzene, C6H6 (l), to gaseous acetylene, C2H2 (g):C6H6 (l) → 3 C2H2 (g)         ΔH = +630 kJ(a) What is the enthalpy change for the reverse reaction? 

Calculate the enthalpy change for the reactionP4O6 (s) + 2 O2 (g) → P4O10 (s)given the following enthalpies of reaction:P4 (s) + 3 O2 (g) → P4O6 (s)          ΔH = -1640.1 kJP4 (s) + 5 O2 (g) → P4O10 (s)        ΔH = -2940.1 kJ 

From the enthalpies of reaction2 H2(g) + O2(g) → 2 H2O(g)          Δ = -483.6 kJ             3 O2(g) → 2 O3(g)            Δ = +284.6 kJcalculate the heat of the reaction3 H2(g) + O3(g) → 3 H2O(g)

From the enthalpies of reactionH2 (g) + F2 (g) → 2 HF (g)          ΔH = -537 kJC (s) + 2 F2 (g) → CF4 (g)          ΔH = -680 kJ2 C (s) + 2 H2 (g) → C2H4 (g)     ΔH = +52.3 kJcalculate ΔH for the reaction of ethylene with F 2:C2H4 (g) + 6 F2 (g) → 2 CF4 (g) + 4 HF (g) 

Given the dataN2 (g) + O2 (g) → 2 NO (g)            ΔH = +180.7 kJ2 NO (g) + O2 (g) → 2 NO2 (g)      ΔH = -113.1 kJ2 N2O (g) → 2 N2 (g) + O2 (g)       ΔH = -163.2 kJuse Hess’s law to calculate ΔH for the reactionN2O (g) + NO2 (g) → 3 NO (g)

Use the following equationsC (s) + O2 (g) → CO2 (g)                                                  ΔH     o = 393.5 kJH2 (g) + 1/2 O2 (g) → H2O (l)                                           ΔH     o = 285.8 kJ2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (l)           ΔH o = 5754.6 kJto calculate the heat formation, ΔH of, for butane.4 C (s) + 5 H2 (g)  →  C4H10 (g)A.  125.7 kJB. -5880.3 kJC. -5862.3 kJD. -5075.3 kJE. -251.4 kJ  

Coal gasification can be represented by the equation:2 C(s) + 2 H2O(g) → CH4(g) + CO2(g)          ΔH = ?Use the following information to find ΔH for the reaction above.CO(g) + H2(g) → C(s) + H2O(g)                    ΔH = −131 kJCO(g) + H2O(g) → CO2(g) + H2(g)               ΔH = −41 kJCO(g) + 3 H2(g) → CH4(g) + H2O(g)            ΔH = −206 kJ      A) −15 kJ            B) 15 kJ     C) −372 kJ       D) 116 kJ          E) −116 kJ 

S(s) + O2(g) → SO2 (g)      ∆H = ‐296.8 kJ2SO2 + O2 → 2SO3             ∆H = ‐198.4 kJWrite the equation for the formation of sulfur trioxide gas from sulfur solid and oxygen. What is the ∆H of that equation?  

Calculate ΔH for the reaction C4H4(g) + 2H2(g) → C4H8(g), using the following information:ΔHcombustion for C4H4(g)= -2341 kJ/molΔHcombustion for C4H8(g)= -3071 kJ/molΔHcombustion for H2(g) = -286 kJ/mola. -128 kJb. -316 kJc. +158 kJd. -700 kJe. -4810 kJ

Calculate ΔH°rxn for the reaction2 Ni(s) + 2 S(s) + 3 O2(g) → 2 NiSO3(s)from the following information:(1) NiSO3(s) → NiO(s) + SO2(g)       ΔH°rxn = 156 kJ (2) S(s) + O2(g) → SO2(g)                 ΔH°rxn = -297 kJ  3) Ni(s) + 1/2 O2(g) → NiO(s)           ΔH°rxn = -241 kJ 

H2(g) + F2(g) → 2HF (g)           ΔH rxn = -546.6 kJ2H2(g) + O2(g) → 2H2O (l)       ΔH rxn = -571.6 kJCalculate the value of  ΔHrxn for:2F2(g) + 2H2O(l) → 4HF(g) + O2(g)

Calculate the enthalpy of the reaction4B(s) + 3O2(g) → 2B2O3(s)given the following pertinent information:B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g),      ΔH∘A=+2035 kJ2B(s) + 3H2(g) → B2H6(g),                            ΔH∘B=+36 kJH2(g) + 1/2 O2(g) → H2O(l),                            Δ  H∘C=−285 kJH2O(l) → H2O(g),                                          ΔH∘D=+44 kJExpress your answer with the appropriate units.

Calculate the enthalpy of the reaction2NO(g) + O2(g) → 2NO2(g)given the following reactions and enthalpies of formation:12N2(g) + O2(g) → NO2(g),   ΔH∘A = 33.2 kJ12N2(g) + 1/2 O2(g) → NO(g),  ΔH∘B = 90.2 kJ

Calculate the enthalpy of the reaction4B(s) + 3O2(g) → 2B2O3(s)given the following pertinent information:A. B2O3(s) + 3H2O(g) → 3O2(g) + B2H6(g),       ΔH∘A = +2035 kJB. 2B(s) + 3H2(g) → B2H6(g),   ΔH∘B = +36 kJC. H2(g) + 1/2O2(g) → H2O(l),    ΔH∘C = −285 kJD. H2O(l) → H2O(g),     ΔH∘D = +44 kJ

Now consider the following set of reactions:2NO2 → 2NO + O2, H=109 kJ/molN2O4 → 2NO + O2, H=172 kJ/molThe equations given in the problem introduction can be added together to give the following reaction:overall: N2O4 → 2NO2 However, one of them must be reversed. Which one:a. 2NO2 → 2NO + O2b. N2O4 → 2NO + O2

Consider these reactions, where M represents a generic metal. Use the information below to determine the enthalpy of the following reaction. 2M(s) + 3Cl2(g) → 2MCl3(s) 

Consider the following set of reactions:CO + 1/2O2 → CO2              ΔH = −283 kJ/molC + O2 → CO2                    ΔH = −393 kJ/molThe equations given in the problem can be added together to give the following reaction:overall: C + 1/2O2 → CO Which one of them must be reversed?

Using the enthalpy of reaction for the 2 reactions with ozone, determine the enthalpy of reaction for the reaction of chlorine with ozone. ClO(g) + O3(g) → Cl(g) + 2O2(g)       ΔH°rxn =  -122.8 kJ2O3(g) → 3O2(g)                               ΔH°rxn =  -285.3 kJO3(g) + Cl(g) → ClO(g) + O2(g)         ΔH°rxn = ?

Which of the following is a statement of Hess's law?a. if a reaction carried out in a series of steps the, ΔH for the reaction will equal the sum of the enthalpy changes for the individual steps.b. if a reaction carried out in a series of steps the, ΔH for the reaction will equal the product of the enthalpy changes for the individual steps.c. the ΔH for a process in the forward direction is equal to the ΔH for the process in the reverse direction.d. the ΔH for a process in the forward direction is equal in magnitude and opposite in sign to the ΔH for a process in the reverse direction.e. the ΔH of a reaction depend on the physical state of the reactants and products.

Calculate ΔHrxn for the following reaction: Express the enthalpy in kilojoules to one decimal place.CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)given these reactions and their ΔH values:C(s) + 2H2(g) → CH4(g)    ΔH = −74.6 kJC(s) + 2Cl2(g) →CCl4(g)   ΔH = −95.7 kJH2(g) + Cl2(g) →2HCl(g)  ΔH = −184.6 kJ

Consider these reactions, where M represents a generic metal1. 2M(s) + 6HCl(aq) → 2MCl3(aq) + 3H2(g) ΔH1 = -749.0 KJ2. HCl(g) → HCl(aq) ΔH2 = -74.8 KJ3. H2(g)+Cl2(g) → 2HCl(g) ΔH3 = -1845.0 KJ4. MCl3(s) → MCl3(aq) ΔH4 = -298.0 KJUse the information above to determine the enthalpy of the following reaction.2M(s) + 3Cl2(g) → 2MCl3(s) ΔH = 

Using Hess' Law, calculate the enthalpy of formation for C 3H8(g) using the following thermodynamic data.C(s) + O2(g) → CO2(g) ΔH = -393.5 kj/molH2(g) + 1/2O2(g) → H2O(I) ΔH = -285.8 kj/molC3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(I) ΔH = 2199.0 kJ/mol

Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25°C, diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Using equations from the list below, determine ΔH  for C(diamond) ⟶ C(graphite)(1) C(diamond) + O2(g) ⟶ CO2(g)          ΔH = −395.4 kJ(2) 2CO2(g) ⟶ 2CO(g) + O2(g)              Δ H = 566.0 kJ(3) C(graphite) + O2(g) ⟶ CO2(g)          Δ H = −393.5 kJ(4) 2CO(g) ⟶ C(graphite) + CO2(g)      ΔH = −172.5 kJ

Given the following data2 O3 (g) → 3 O2 (g)            ΔH = -427 kJO2 (g) → 2O (g)                  ΔH = 495 kJNO (g) + O3 (g) → NO2 (g) + O2 (g)   ΔH = -199 kJcalculate ΔH for the reactionNO (g) + O (g) → NO2 (g) 

Consider the following balanced thermochemical equation for the decomposition of the mineral magnesite:MgCO3 (s) ⟶ MgO (s) + CO2 (g)                         ΔH = 117.3 kJWhat is ΔH for the reverse reaction?

Find ΔHrxn for the following reaction: N2O(g) + NO2(g) → 3 NO(g)Use the following reactions with known ΔH values:2 NO(g) + O2(g) → 2 NO2(g); ΔH = –113.1 kJN2(g) + O2(g) → 2 NO(g); ΔH = +182.6 kJ2 N2O(g) → 2 N2(g) + O2(g); ΔH = –163.2 kJ

Consider the following set of reactions:N2 + 2O2 → N2O4 ,ΔH=−8 kJ/molN2 + O2 → 2NO ,ΔH=180 kJ/molThe equations given in the problem introduction can be added together to give the following reaction:overall: N2O4 → 2NO + O2However, one of them must be reversed, reaction 1: N2 + 2O2 → N2O4What is the enthalpy for reaction 1 reversed?reaction 1 reversed: N2O4 → N2 + 2O2Express your answer numerically in kilojoules per mole.What is the enthalpy for reaction 2?

Consider these reactions, where M represents a generic metal.1.  2M (s) + 6HCl (aq)  →  2MCl3 (aq) + 3H2 (g)                 ΔH1 = -579.0 kJ2. HCl (g) → HCl (aq)                                                         ΔH2 = -74.8 kJ3. H2 (g) + Cl2 (g) → 2HCl (g)                                            ΔH3 = -1845.0 kJ4. MCl3 (s) → MCl3 (aq)                                                     ΔH4 = -141.0 kJUse the following information above to determine the enthalpy (kJ) of the following reaction.2 M(s) +  3Cl2 (g) → 2 MCl3 (s)

Calculate ΔHrxn for the reaction CaO(s) + CO2(g) → CaCO3(s) given these reactions and their ΔH values: Express the enthalpy in kilojoules to one decimal place.Ca(s) + CO2(g) + 1/2O2 → CaCO3 (s)   ΔH = −814.1 kJ2Ca(s) + O2(g) → 2CaO(s)                    ΔH = −1269.8 kJ

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: 2S(s) + 3O2(g) → 2SO3(g)     ΔH°rxn = ? Given: SO2(g) → S(s) + O2(g)       ΔH°rxn = +296.8 kJ 2SO2(g) + O2(g) → 2SO3(g)       ΔH°rxn = -197.8 kJ a. -494.6 kJ b. -692.4 kJ c. -791.4 kJ d. -98.8 kJ e. -293.0 kJ

The ΔHf˚ of TiI3(s) is –328 kJ/mol and the ΔH˚ for the reaction Ti(s) + 3 I 2(g) → 2 TiI3(s) for the reaction is –839 kJ{ m kJ}. Calculate the ΔH of sublimation of I2(s), which is a solid at 25˚C.

A chemist measures the enthalpy change ΔH during the following reaction: 2H2O (l) + O2 (g) → 2H2O2 (l)            ΔH = 196. kJ Use this information to calculate ΔH of each reaction. Round each of your answers to the nearest kJ/mol.

Given the following reactions:(1) 2NO --> N2 + O2            ∆H = -180 kJ(2) 2NO + O2 --> 2NO2      ∆H = -112 kJthe enthalpy of the reaction of nitrogen with oxygen to produce nitrogen dioxide2N2 + 4O2 --> 4NO2is _________.a. -136 kJb. 68kJc. 136 kJd. 292kJe. -292 kJ

Consider these reactions, where M represents a generic metal2M(s) + 6HCl(aq) → 2MCl3(aq) + 3H2(g)              H = -748.0 kJHCl(g) → HCl(aq)                                                   H = -74.8H2(g) + Cl2(g) → 2HCl(g)                                       H = -1845MCl3(s) → MCl3(aq)                                               H = -317.0Using the information above to determine the enthalpy of the following reaction2M(s) + 3Cl2(g) → 2MCl3(s)

Calculate the enthalpy of formation of SO2 (g) from the standard enthalpy changes of the following reactions: 2SO2 (g) + O2 (g) → 2SO3(g)             ΔH°rxn = -196 kJ 2S (s) + 3O2 (g) → 2SO3 (g)              ΔH°rxn = -790 kJ S (s) + O2 (g) → SO2 (g)                    ΔH°rxn = ?

Use the following data to determine the Delta H for the conversion of diamond into graphite: Cdiamond (s) + O2 (g) → CO2 (g)    ΔH° = -395.4 kJ 2CO2 (g) → 2CO (g) + O2 (g)     ΔH° = 566.0 kJ 2CO (g) → Cgraphite (s) + CO2 (g)     ΔH° = -172.5 kJ Cdiamond (s) → Cgraphite (s)     ΔH° = ?

Consider the equations below. (1) Ca (s) + CO2 (g) + 1/2O2 (g) → CaCO3 (s) (2) 2Ca (s) + O2 (g) → 2CaO (s) How should you manipulate these equations so that they produce the equation below when added: Check all that apply. CaO (s) + CO2 (g) → CaCO3 (s) (A) reverse the direction of equation (2) (B) multiply equation (1) by 3 (C) multiply equation (2) by 1/2

Given these reactions, X (s) + 1/2O2 (g) → XO (s)     ΔH = -993.7 kJ XCO3 (s) → XO (s) + CO2 (g)     ΔH = +424.5 kJ What is ΔHrxn for this reaction? X (s) + 1/2O2 (g) + CO2 (g) → XCO3 (s) 

Calculate Δ Hrxn for the following reaction:CaO(s) + CO2 (g) ---> CaCO3 (s)Use the following reactions and given Δ H values:Ca(s) + CO2(g) + 1/2 O2(g) --> CaCO3 (s), ΔH = -812.8 kJ2Ca(s) + O2 (g) --> 2CaO(s),  ΔH = -1269.8 kJExpress your answer using four significant figures.

Research is being carried out on cellulose as a source of chemicals for the production of fibers, coatings and plastics. Cellulose consists of long chains of glucose molecules (C6H12O6), so for the purposes of modeling the reaction we can consider the conversion of glucose to formaldehyde (H2CO). Calculate the heat of reaction for the conversion of 1 mole of glucose into formaldehyde, given the following thermochemical data:H2CO (g) + O2 (g) → CO2 (g) + H2O (g)     ΔH°comb = -572.9 kJ/mol 6C (s) + 6H2 (g) + 3O2 (g) → C6H12O6 (s)     ΔH°f = -1274.4 kJ/mol C (s) + O2 (g) → CO2 (g)     ΔH°f = -393.5 kJ/mol H2 (g) + 1/2O2 (g) → H2O (g)     ΔH°f = -285.8 kJ/mol C6H12O6 (s) → 6H2CO (g)     ΔH°rxn = ?

Find ΔH° for BaCO3 (s) → BaO (s) + CO2 (g) given that 2Ba (s) + O2 (g) → 2BaO (s)     ΔH° = -1107.0 kJBa (s) + CO2 (g) + 1/2O2 (g)→ BaCO3 (s)     ΔH° = - 822.5 kJ a. -1929.5 kJ b. -1376.0 kJ c. -284.5 kJ d. 269.0 kJ e. 537 kJ

From the enthalpies of reaction 2C(s) + O2(g) → 2CO(g)                             ΔH = -221.0 kJ 2C(s) + O2(g) + 4H2(g) → 2CH3OH(g)       ΔH = -402.4kJ calculate ΔH for the reaction:CO(g) + 2H2 → CH3OH(g) Express your answer with the appropriate units.

Given that H2 (g) + F2 (g) → 2HF (g)     ΔH°rxn = - 546.6 kJ 2H2 (g) + O2 (g) → 2H2O (l)     ΔH°rxn = - 571.6 kJ calculate the value of ΔH°rxn for 2F2 (g) + 2H2O (l) → 4HF (g) + O2 (g) 

Calculate the enthalpy of the reaction 2NO (g) + O2 (g) → 2NO2 (g) given the following reactions and enthalpies of formation: 1. 1/2N2 (g) + O2 (g) → NO2 (g), ΔH°A  = 33.2 kJ 2. 1/2N2 (g) + 1/2O2 (g) → NO (g), ΔH°B = 90.2 kJ Calculate the enthalpy of the reaction.Express your answer with the appropriate units.

The combustion reaction of butane is as follows. C4H10(g) + 13/2 O2(g) → 4 CO2(g) + 5 H2O(l) Using Hess's law and the reaction enthalpies given below, find the change in enthalpy for this reaction. reaction (1):    C(s) + O2(g) → CO2(g)                               ΔH = -393.5 kJ/mol reaction (2):   H2(g) + 1/2 O2(g) → H2O(I)                        ΔH = -285.8 kJ/mol reaction (3):   4C(s) + 5H2(g) → C4H10 (g)                       ΔH = -125.7 kJ/mol

Given the following data: N2(g) + O2(g) → 2NO(g),          ΔH = +180.7 kJ 2NO(g) + O2(g) → 2NO2(g),     ΔH = -113.1 kJ 2N2O(g) → 2N2(g) + O2(g),      ΔH = -163.2 kJ Use Hess's law to calculate ΔH for the following reaction: N2O(g) + NO2(g) → 3NO(g) Express the enthalpy in kilojoules to four significant digits. 

Consider the following chemical reaction. NH3(g) + 2 O2(g) → HNO3(aq) + H2O(l) Calculate the change in enthalpy (ΔH) for this reaction, using Hess' law and the enthalpy changes for the reactions given below.

The enthalpy of decomposition of NO2Cl is -114 kJ. Use the following data to calculate the heat of formation of NO2Cl from N2, O2, and Cl2: NO2Cl (g) → NO2 (g) + 1/2Cl2(g)    ΔH°rxn = -114 kJ 1/2 1/2N2 (g) + O2 (g) → NO2 (g)    ΔH°f = 33.2 kJ 1/2 N2 (g) + O2 (g) + 1/2 Cl2 (g) → NO2Cl (g)    ΔH°f  = ?

Use the following data to calculate the standard heat (enthalpy) of formation, ΔH °f, of manganese(IV) oxide, MnO2 (s). 2MnO2 (s) → 2MnO (s) + O2 (g)          ΔH = 264 kJ MnO2 (s) + Mn(s) → 2MnO (s)           ΔH = -240 kJ Select one:(a) -504 kJ (b) -372 kJ (c) 24 kJ (d) -24 kJ (e) 504 kJ

Calculate ΔHrxn for the following reaction:CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)Use the following reactions and given ΔH′s.C(s) + 2H2(g) → CH4(g)     ΔH = −74.6kJC(s) + 2Cl2(g) → CCl4(g)    ΔH = −95.7kJH2(g) + Cl2(g) → 2HCl(g)   ΔH = −184.6kJ