| Sections | |||
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| Internal Energy | 22 mins | 0 completed | Learn |
| Calorimetry | 39 mins | 0 completed | Learn |
| Thermochemical Equation | 6 mins | 0 completed | Learn Summary |
| Hess's Law | 17 mins | 0 completed | Learn |
| Enthalpy of Formation | 13 mins | 0 completed | Learn Summary |
| End of Chapter 6 Problems | 52 mins | 0 completed | Learn |
| Additional Practice |
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| Units of Energy |
| Additional Guides |
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| Endothermic & Exothermic Reactions |
| Enthalpy |
Many chemical reactions require multiple steps in order to create their final products. Hess’s Law states that determining the enthalpy or heat of reaction (under constant pressure) is independent of the pathway taken to make those final products.
Understanding Hess’s Law
Hess’s Law states that the enthalpy of reaction (ΔHRxn) is the sum of the enthalpy changes of its individual thermochemical steps and represents a state function.
For example, the oxidation of acetylene is represented by the following equation:
Thermochemical Equation (Oxidation of Acetylene)
In order to determine its enthalpy of reaction we use the enthalpies of its individual thermochemical steps.
Thermochemical Steps (Oxidation of Acetylene)
Since each thermochemical step balances out to give the final equation we can add together the enthalpy values of ΔH1, ΔH2 and ΔH3 to obtain the ΔHRxn.
Total Enthalpy Change
Application of Hess’s Law
There will be occasions when the individual thermochemical steps will not balance out to give the final overall equation. In these cases we must manipulate and alter them.
PRACTICE 1: Find the ΔHRxn for the given overall chemical equation:
Thermochemical Step (Xenon Tetrafluoride)
The given thermochemical steps and ΔH values are
Thermochemical Steps (Xenon Tetrafluoride)
STEP 1: Start with XeF2 from the overall chemical equation and locate where it is in the given thermochemical steps. If XeF2 does not match the XeF2 in the overall chemical equation then we will need to manipulate the thermochemical step.
Application of Hess's Law (XeF2)
In our thermochemical step XeF2 is listed as a product, but in our overall chemical equation XeF2 is a reactant. This means we must reverse the thermochemical step so that XeF2 will now be a reactant.
Reversing a thermochemical step
By reversing the thermochemical step we have XeF2 as a reactant just like in the overall chemical equation. Also notice that reversing a thermochemical step causes a reversal of the sign for its ΔH value.
STEP 2: Continue to the next compound in the overall chemical equation, F2, and locate where it is in the given thermochemical steps. As we did before, if F2
does not match the F2 in the overall chemical equation then we will need to manipulate the thermochemical step.
Thermochemical Steps (F2)
We find that F2 is found in both thermochemical steps. Whenever a compound is found in more than one step then we ignore it for now and move onto our next compound in the overall chemical equation.
STEP 3: Next we look at XeF4 from the overall chemical equation and locate where it is in the given thermochemical steps. If necessary, manipulate the thermochemical step.
Thermochemical Step (XeF4)
The XeF4 in the thermochemical step matches with the XeF4 found in the overall chemical equation. This means we don’t have to do anything in this step.
STEP 4: Now bring down all the thermochemical steps and cancel out the reaction intermediates to obtain the overall chemical equation.
Canceling Reaction Intermediates
STEP 5: Canceling out the intermediates gives us the overall chemical equation and by adding the two ΔHo values of the thermochemical steps you can isolate the ΔHRxn.
Calculation of Enthalpy of Reaction
Now test your skills with a more complex Hess’s Law problem.
PRACTICE 2: Find the ΔHRxn for the given overall chemical reaction:
Thermochemical Equation (Carbon Dioxide)
The given thermochemical steps and ΔH values are
Thermochemical Steps (Carbon Dioxide)
STEP 1: Start with CO2 from the overall chemical equation and locate where it is in the given thermochemical steps. If CO2 does not match the CO2 in the overall chemical equation then we will need to manipulate the thermochemical step.
Thermochemical Step (Carbon Dioxide)In our thermochemical step CO2 is listed as a reactant like we need, but there are two moles of it. In the overall chemical equation we have only 1 mole of CO2 so we must divide the thermochemical equation by 2.
Reversal of a Thermochemical Step
Realize that dividing the thermochemical step by 2 also means you divide your original ΔH value of 3511.1 kJ by 2. Dividing 3511.1 kJ by 2 gives a new ΔH value of 1755.55 kJ.
STEP 2: Continue to the next compound in the overall chemical equation, C (s), and locate where it is in the given thermochemical steps. If necessary, manipulate the thermochemical step.
Thermochemical Step (Carbon solid)
In our thermochemical step we have two moles of C (s) listed products, but we need only 1 mole. So we divide the thermochemical step by 2.
Thermochemical Step (Dividing by 2)
Dividing by 2 means we also divide the ΔH value by 2.
STEP 3: Next we search for O2 in the given thermochemical steps. If necessary, manipulate the thermochemical step.
Thermochemical Steps (Oxygen Gas)We find O2 in more than one thermochemical step so we can ignore it for it.
STEP 4: Now bring down all the thermochemical steps and cancel out the reaction intermediates to obtain the overall chemical equation.
Thermochemical Steps (Carbon Dioxide)We must cancel out H2O and H2 because they are not found in the overall chemical equation. In order to cancel out the H2 and H2O molecules we will need to reverse and multiple the first thermochemical step by 1.5 or 3/2.
Thermochemical Step (Reversal & Multiplication)
Reversing the thermochemical step reverses the sign for the ΔH value and multiplying the thermochemical step by 1.5, which is the same as 3/2, means we multiply the ΔH value by 1.5. Now we bring down all the thermochemical steps again and finally cancel out all reaction intermediates. 
Canceling Reaction Intermediates (Carbon Dioxide)
STEP 5: Canceling out the intermediates gives us the overall chemical equation and by adding the three ΔHo values of the thermochemical steps you can isolate the ΔHRxn.
Overall Enthalpy Change
Hess’s Law serves as one component in our understanding of the branch of chemistry dealing with thermal and chemical energy called Thermochemistry. Under Thermochemistry, we examine the concepts of calorimetry, the conservation of energy, internal energy, formation equations, thermochemical equations and stoichiometry.
Additional Problems
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
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 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
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 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
S(s) + O2(g) → SO2 (g) ∆H = ‐296.8 kJ
2SO2 + O2 → 2SO3 ∆H = ‐198.4 kJ
Write the equation for the formation of sulfur trioxide gas from sulfur solid and oxygen. What is the ∆H of that equation?
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.
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 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
Use the following equations
C (s) + O2 (g) → CO2 (g) ΔH o = 393.5 kJ
H2 (g) + 1/2 O2 (g) → H2O (l) ΔH o = 285.8 kJ
2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (l) ΔH o = 5754.6 kJ
to calculate the heat formation, ΔH of, for butane.
4 C (s) + 5 H2 (g) → C4H10 (g)
A. 125.7 kJ
B. -5880.3 kJ
C. -5862.3 kJ
D. -5075.3 kJ
E. -251.4 kJ
Determine the answer for the question below.
The heat of solution is LiCl is –37.1 kJ/mol, and the lattice energy of LiCl(s) is 828 kJ/mol. Calculate the total heat of hydration of 1 mol of gas phase Li+ ions and Cl - ions.1) 791 kJ2) 865 kJ3) -865 kJ4) -791 kJ5) None of these
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?
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.
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 = ?
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 =
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 metal. Use the information below to determine the enthalpy of the following reaction. 2M(s) + 3Cl2(g) → 2MCl3(s)
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
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.
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 ΔHrxn for:
2F2(g) + 2H2O(l) → 4HF(g) + O2(g)
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
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)
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
Calculate ΔH°rxn for the reaction
2 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
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
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
At 1 atm pressure, the heat of sublimation of gallium is 277 kJ/mol and the heat of vaporization is 271 kJ/mol. To the correct number of significant figures, how much heat is required to melt 4.50 mol of gallium at 1 atm pressure?a. 250 kJb. 6 kJc. 274 kJd. 27 kJ
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
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
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)
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
Consider the following generic reaction:
A+2B→C+3D, with ΔH = 103 kJ .
Determine the value of ΔH for each of the following related reactions.
a. 3A + 6B → 3C + 9D
b. C + 3D → A + 2B
c. 12C + 32D → 12A + B
Calculate the ΔHrnx for the following reaction C(s) + H2O(g) → CO(g) + H2(g)Use the following reactions and given ΔH'sC(s) + O2(g) → CO2(g) Δh = -393.5 kJ2CO(s) + O2(g) → 2CO2(g) Δh = -566.0 kJ2H2(g) + O2(g) → 2H2O(g) Δh = -483.6 kJ
Even though so much energy is required to form a metal cation with a 2+ charge, the alkaline earth metals form halides with the general formula MX2, rather than MX.(c) Use Hess’s law to calculate ΔH° for the conversion of MgCl to MgCl2 and Mg (ΔH° f of MgCl2 = −641.6 kJ/mol).
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
Calculate ΔH for the process Hg 2Cl2(s) ⟶ 2Hg(l) + Cl2(g) from the following information:Hg(l) + Cl2(g) ⟶ HgCl2(s) ΔH = −224 kJHg(l) + HgCl2(s) ⟶ Hg2Cl2(s) ΔH = −41.2 kJ
Calculate ΔH°298 for the process Co3O4(s) ⟶ 3Co(s) + 2O2(g) from the following information:Co(s) + 1/2O2(g) ⟶ CoO(s) ΔH°298 = −237.9kJ 3CoO(s) +1/2O2(g) ⟶ Co3O4(s) ΔH°298 = −177.5kJ
Calculate the standard molar enthalpy of formation of NO(g) from the following data:N2(g) + 2O2 ⟶ 2NO2(g) ΔH°298 = 66.4 kJ2NO(g) + O2 ⟶ 2NO2(g) ΔH°298 = −114.1 kJ
Calculate ΔH°298 for the process Zn(s) + S(s) + 2O 2(g) ⟶ ZnSO4(s)from the following information:Zn(s) + S(s) ⟶ ZnS(s) ΔH° 298 = −206.0 kJZnS(s) + 2O2(g) ⟶ ZnSO4(s) ΔH°298 = −776.8 kJ
Given the following dataFe2O3 (s) + 3 CO (g) → 2 Fe (s) + 3 CO2 (g) ΔH° = -23 kJ3Fe2O3 (s) + CO (g) → 2 Fe3O4 (s) + CO2 (g) ΔH° = -39 kJFe2O3 (s) + CO (g) → 3FeO (s) + CO2 (g) ΔH° = 18 kJcalculate ΔH° for the reactionFeO (s) + CO (g) → Fe(s) + CO2 (g)
For each generic reaction, determine the value of ΔH2 in terms of ΔH1.A + B → 2C; ΔH12C → A + B; ΔH2 = ?
For each generic reaction, determine the value of ΔH2 in terms of ΔH1.A → B + 2C; ΔH11/2B + C → 1/2A; ΔH2 = ?
Calculate ΔHrxn for the following reaction: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)Use the following reactions and given ΔH's.2 Fe(s) + 3/2 O2(g) → Fe2O3(s), ΔH = –824.2 kJCO(g) + 1/2 O2(g) → CO2(g), ΔH = –282.7 kJ
Calculate ΔHrxn for the following reaction: 5 C(s) + 6 H2(g) → C5H12(l)Use the following reactions and given ΔH's.C5H12(l) + 8 O2(g) → 5 CO2(g) + 6 H2O(g), ΔH = –3244.8 kJC(s) + O2(g) → CO2(g), ΔH = –393.5 kJ2 H2(g) + O2(g) → 2 H2O(g), ΔH = –483.5 kJ
Given the following data:2ClF (g) + O2 (g) → Cl2O (g)+ F2O (g) ΔH = 167.4 kJ2ClF3 (g) + 2O2 (g) → Cl2O (g) + 3F2O (g) ΔH = 341.4 kJ2F2 (g) + O2 (g) → 2F2O (g) ΔH = -43.4 kJcalculate ΔH for the reactionClF (g) + F2 (g) → ClF3 (g)
Calculate ΔH for the reaction2NH3 (g) + 1/2 O2 (g) → N2H4 (l) + H2O (l) given the following data:2NH3 (g) + 3N2O (g) → 4N2 (g) + 3H2O (l) ΔH = -1010 kJN2O (g) + 3H2 (g) → N2H4 (l) + H2O (l) ΔH = -317 kJN2H4(l) + O2 (g) → N2(g) + 2H2O (l) ΔH = -623 kJH2(g) + 1/2O2 (g) → H2O (l) ΔH = -286 kJ
Calculate the enthalpy change for the reaction: P4O6(s) + 2 O2(g) → P4O10(s), given the following enthalpies of reaction:P4(s) + 3 O2(g) → P4O6(s), ΔH = –1640.1 kJP4(s) + 5O2(g) → P4O10(s), ΔH = –2940.1 kJ
Calculate ΔH forCa (s) + 1/2O2 (g) + CO2 (g) ⟶ CaCO3 (s)given the following reactions:Ca (s) + 1/2O2 (g) ⟶ CaO (s) Δ H = −635.1 kJCaCO3 (s) ⟶ CaO (s) + CO2 (g) ΔH = 178.3 kJ
Calculate ΔH for2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)given the following reactions:1/2N2 (g) + 1/2O2 (g) ⟶ NO (g) Δ H = 90.3 kJNO (g) + 1/2Cl2 (g) ⟶ NOCl (g) Δ H = −38.6 kJ
Write the balanced overall equation (equation 3) for the following process, calculate ΔHoverall, and match the number of each equation with the letter of the appropriate arrow in Figure P6.72:(1) N2 (g) + O2 (g) ⟶ 2NO (g) Δ H = 180.6 kJ(2) 2NO (g) + O2 (g) ⟶ 2NO2 (g) Δ H = −114.2 kJ--------------------------------------------------------------------------------------------(3) ΔH overall = ?
Calculate ΔHrxn for the following reaction: CaO(s) + CO2(g) → CaCO3(s)Use the following reactions and given ΔH’s:Ca(s) + CO2(g) + 1/2 O2(g) → CaCO3(s); ΔH = –812.8 kJ2 Ca(s) + O2(g) → 2 CaO(s); ΔH = –1269.8 kJ
Calculate ΔHrxn Delta H_{
m rxn} for the following reaction: CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)Given these reactions and their ΔH values:C(s) + 2 H2(g) → CH4(g); ΔH = –74.6 kJC(s) + 2 Cl2(g) → CCl4(g); ΔH = –95.7 kJH2(g) + Cl2(g) → 2 HCl(g); ΔH = –92.3 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.
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?
Given the following thermochemical data:Fe2O3(s) + CO(g) → 2 FeO(s) + CO2(g) ∆H = −2.9 kJFe(s) + CO2(g) → FeO(s) + CO(g) ∆H = 11.3 kJdetermine ∆H for the following reaction:Fe2O3(s) + 3 CO(g) −−→ 2 Fe(s) + 3 CO2(g)(A) 8.4 kJ (B) -8.4 kJ (C) -25.5 kJ (D) 25.5 kJ (E) -19.7 kJ
From the enthalpies of reaction2 C(s) + O2(g) → 2CO(g) ΔH = -221.0 kJ2C(s) + O2(g) + 4H2(g) → 2CH3OH(g) ΔH = -402.4 kJcalculate ΔH for the reaction:CH3OH(g) → CO(g) + 2H2(g) a) -90.7 kJ b) 90.7 kJ c) -623.4 kJ d) 180.7 kJ e) 623.4 kJ
Given the following thermochemical data: ∆H°f (kJ)I. P4(s) + 6 Cl2(g) ⟶ 4 PCl3(g) –1225.6II. P4(s) + 5 O2(g) ⟶ P4O10(s) –2967.3III. PCl3(g) + Cl2(g) ⟶ PCl5(g) –84.2IV. PCl3(g) + 1/2 O2(g) ⟶ POCl3(g) –285.7Calculate the value of ∆H° for the following reaction:P4O10(s) + 6 PCl5(g) ⟶ 10 POCl3(g)A. –110.5 kJ B. –610.1 kJ C. –2682.2 kJ D. –4562.8 kJE. –7555.0 kJ
(i) Consider this reactionC2H5OH + 3O2 → 2CO2 + 3H2O Change in Heat = -1370 kJ/molWhat is the enthalpy for this reaction reversed?Reaction reversed: 2CO2 + 3H2O → C2H5OH + 3O2Change in Heat = ? kJ/mol(ii) Calculate the enthalpy of the reaction2NO(g) + O2(g) → 2NO2(g)Given the following reactions and enthalpies of formation:½ N2(g) + O2(g) → NO2(g) Change in Heat a = 33.2 KJ½ N2(g) + ½ O2 (g) → NO(g) Change in Heat b = 90.2 KJExpress your answer numerically in kilojoules.Change in Heat = ? kJ/molPlease show all your steps and explain why you did what.
Match each chemical reaction with the correct diagram according to Hess' LawΔHC = -ΔHA + ΔHB ΔHB = ΔHA + ΔHCΔHA = ΔHB - ΔHCΔHB = -ΔHA - ΔHC ΔHA = -ΔHB + ΔHCΔHC = ΔHA + ΔHB ΔHB = ΔHA - ΔHC ΔHC = -ΔHA - ΔHB
Enthalpy diagram illustrating Hesss law. The net reaction is the same as in Figure 5.21 in the textbook, but here we imagine different reactions in our two-step version. As long as we can write a series of equations that add up to the equation we need, and as long as we know a value for H for all intermediate reactions, we can calculate the overall H.Suppose the overall reaction were modified to produce 2H2O(g) rather than 2H2O(l). Would any of the values of H in the diagram stay the same?
Given:1. A → 2B, ΔH1 2. B → C + D, ΔH2 3. E → 2D, ΔH3ΔH of A → 2C + E is:A. ΔH1 + 2ΔH2 – ΔH3 B. ΔH2 - ΔH2 – ΔH3 C. ΔH3 – ΔH3 D. ΔH1 + 2ΔH1 – ΔH2 E. None of the above
Calculate the ΔH (in kJ) for the reaction:3 Fe2O3 (s) + CO (g) → CO2 (g) + 2 Fe3O4 (s) ∆Hrxn = ?Given the following reactions:Fe2O3 (s) + 3 CO (g) → 2 Fe (s) + 3 CO2 (g) ∆H1 = −28. 0 kJ3 Fe (s) + 4 CO2 (g) → 4 CO (g) + Fe3O4 (s) ∆H2 = +12. 5 kJA. -59.0B. +40.5C. -15.5D. -109E. +109
Given the following reactionsH2O (l) →H2O (g) ΔH = 44.01 kJ2H2 (g) + O2 (g) → 2H2O (g) ΔH = -483.64 kJthe enthalpy for the decomposition of liquid water into gaseous hydrogeand oxygen2H2O (l) → 2H2 (g) + O2 (g) is __________ kJ.A) -395.62B) -527.65C) 439.63D) 571.66E) 527.65
From the enthalpies of reaction2C(s) + O2 (g) → 2CO(g) ΔH rxn = -221.0 kJ2C(s) + O2 (g) + 4H2 (g) → 2CH3OH(g) ΔHrxn = -402.4 kJcalculate ΔH for the reaction:CO(g) + 2H2 (g) → CH3OH(g)(a) -623.4 kJ (b) 623.4 kJ (c) 181.4 kJ (d) -90.7 kJ (e) -181.4 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?
Calculate ΔH°298 for the process Sb(s) + 5/2Cl2(g) ⟶ SbCl5(s)from the following information:Sb(s) + 3/2Cl2(g) ⟶ SbCl3(s) ΔH°298 = −314 kJSbCl3(s) + Cl2(g) ⟶ SbCl5(g) ΔH°298 = −80 kJ
Calculate the enthalpy of the reaction2NO(g) + O2(g) → 2NO2(g)Given the following reactions and enthalpies of formation:½ N2(g) + O2(g) → NO2(g) Change in Heat a = 33.2 KJ½ N2(g) + ½ O2 (g) → NO(g) Change in Heat b = 90.2 KJExpress your answer numerically in kilojoules.Change in Heat = ? KJ/mol
Calculate ΔH for the reaction N2H4 (l) + O2 (g) → N2 (g) + 2H2O (l)given the following data2 NH3 (g) + 3N2O (g) → 4N2(g) + 3H2O (l) ΔH = -1010 kJN2O (g) + 3 H2 (g) → N2H4 (l) + H2O (l) ΔH = -317 kJ2NH3 (g) + 1/2 O2 (g) → N2H4 (l) + H2O (l) ΔH = -143 kJH2 (g) + 1/2 O2 (g) → H2O (l) ΔH = -286 kJ
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 following reaction2 NOCl (g) → N2 (g) + 02 (g) + Cl2 (g)Given the following set of reactions;1/2 N2 (g) + 1/2 02 (g) → NO (g) ΔH= 90.3KJNO (g) + 1/2 Cl2 (g) → NOCl (g) ΔH = - 38.6 KJ
Given these reactions:X(s) +1/2 O2(g) → XO(s)H=-639.7XCO3(s) → XO(s)+CO2(g)H=+349.7 What is H(rxn) for this reaction?X(s)+1/2 O2(g)+CO2(g) --> XCO3(s)
Given the following dataCa(s) + 2C (graphite) → CaC2 (s) ΔH = -62.8 kJCa(s) + 1/2 O2 (g) → CaO (s) ΔH = -635.5 kJCaO (s) + H2O (l) → Ca(OH)2 (aq) ΔH = -653.1 kJC2H2 (g) + 5/2O2 (g) → 2CO2 (g) + H2O (l) ΔH = -1300. kJC(graphite) + O2 (g) → CO2 (g) ΔH = -393.5 kJCalculate ΔH for the reactionCaC2 (s) + 2H2O (l) → Ca(OH)2 (aq) + C2H2 (g)
Using the following data, calculate the standard heat of formation of ICl(g) in kJ/mol:Cl2 (g) → 2Cl (g) ΔH° = 242.3 kJI2 (g) → 2I (g) ΔH° = 151.0 kJICl (g) → I (g) + Cl (g) ΔH° = 211.3 kJI2 (s) → I2 (g) ΔH° = 62.8 kJ
Consider these reactions, where M represents a generic metal. 1. 2 M(s) + 6 HCl(aq) → 2 MCl3(aq) + 3 H2(g) ΔH1 = -623.0 kJ2. HCl(g) → HCl(aq) ΔH2 = -74.8 kJ3. H2(g) + Cl2(g) → 2 HCl(g) ΔH3 = -1845.0 kJ 4. MCl3(s) → MCl3(aq) ΔH4 = -309.0 kJUse the information above to determine the enthalpy of the following reaction2M(s) + 3Cl2(g) → 2MCl3(s) ΔH = kJ
The following sequence of reactions occurs in the commercial production of aqueous nitric acid:4NH3(g) + 5O2(g) ⟶ 4NO(g) + 6H2O(l) ΔH = −907 kJ2NO(g) + O2(g) ⟶ 2NO2(g) ΔH = −113 kJ3NO2 + H2O(l) ⟶ 2HNO3(aq) + NO(g) ΔH = −139 kJDetermine the total energy change for the production of one mole of aqueous nitric acid by this process.
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
Reaction of gaseous ClF with F 2 yields liquid ClF3, an important fluorinating agent. Use the following thermochemical equations to calculate ΔH°rxn for this reaction:(1) 2CIF(g) + O2(g) ⟶ Cl2O(g) + OF2(g) ΔH°rxn = 167.5 kJ(2) 2F2(g) + O2(g) ⟶ 2OF2(g) ΔH°rxn = −43.5 kJ(3) 2ClF3(l) + 2O2(g) ⟶ Cl2O(g) + 3OF2(g) ΔH°rxn = 394.1 kJ2ClF + 2F2 ⟶ 2ClF3
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 data:Calculate ΔH for the reaction On the basis of the enthalpy change, is this a useful reaction for the synthesis of ammonia?
Calculate the enthalpy change of2A + E -->3Cgiven the following:A + 1/2B --> C + D Delta H = -490.0 kJC + B --> 2D + E Delta H = -160.0 kJ
Use the standard reaction enthalpies given below to determine ∆Hrxn for the following reactionP4 (g) + 10 Cl2(g) ---> 4PCI5(s) ∆Hrxn = ?GivenPCI5(s) --> PCl5(s) + Cl2(g) ∆Hrxn = +157 kJP4(g) + 6 Cl2(g) --> 4 PCl3(g) ∆Hrxn = -1207 kJ-2100. kJ-1050. kJ-1786 kJ-1364 kJ-1835 kJ
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 = ?
A chemist measures the enthalpy change ΔH during the following reaction: CS2 (g) + 3Cl2 (g) → CCl4 (l) + S2Cl2 (l) ΔH = -304. kJ Use this information to complete the table below. Round each of your answers to the nearest kJ/mol.
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)
How can the first two reactions be combined to obtain the third reaction? Include phases in the balanced chemical equation.
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.
Nitroglycerine is a powerful explosive that forms four different gases when detonated:2 C3H5(NO3)3(l) → 3 N2(g) + 1/2 O2(g) + 6 CO2(g) + 5 H2O(g)Calculate the enthalpy change that occurs when 15.0 g of nitroglycerine a detonated. The standard enthalpies of formation are shown below.
Calculate the enthalpy of the reaction 4B (s) + 3O2 (g) → 2B2O3 (s) given the following pertinent information: 1. B2O3 (s) + 3H2O (g) → 3O2 (g) + B2H6 (g), ΔH°A = +2035 kJ 2. 2B (s) + 3H2 (g) → B2H6 (g), ΔH°B = +36kJ 3. H2 (g) + 1/2O2 (g) → H2O (l), ΔH°C = -285 kJ 4. H2O (l) → H2O (g), ΔH°D = +44 kJExpress your answer with the appropriate units.
Given the standard enthalpy changes for the following two reactions: 4C (s) + 5H2 (g) → C4H10 (g) ΔH° = -125.6 kJ C2H4 (g) → 2C (s) + 2H2 (g) ΔH° = -52.3 kJ what is the standard enthalpy change for the reaction: 2C2H4 (g) + H2 (g) → C4H10 (g) ΔH° = ?
A chemist measures the enthalpy change ΔH during the following reaction: P4 (s) + 6Cl2 (g) → 4 PCl3 (g) ΔH = -1148.kJ Use this information to complete the table below. Round each of your answers to the nearest kJ/mol.
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
Enter your answer in the provided box.From these data, S (rhombic) + O2 (g) → SO2 (g) ΔH°rxn = -296.06 kJ/mol S (monoclinic) + O2 (g) → SO2 (g) ΔH°rxn = - 296.36 kJ/mol calculate the enthalpy change for the transformation S (rhombic) → S (monoclinic) (Monoclinic and rhombic are different allotropic forms of elemental sulfur.)
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.
Calculate the ΔH° for the following reaction using the information below it. C2H5OH (l) + O2 (g) → CH3CO2H (l) + H2O (l): ΔH° = ? C2H5OH (l) + 1/2O2 (g) → CH3CHO(l) + H2O(l): ΔH° = -175 kJ CH3CHO (l) + 1/2O2 (g) → CH3CO2H (l): ΔH° = -318 kJ (A) 493 kJ (B) 247 kJ (C) 143 kJ (D) -143 kJ (E) -493 kJ
Using data from the following reactions and applying Hess's law, calculate the heat change for the slow reaction of zinc with water: Zn (s) + 2H2O (l) → Zn2+ (aq) + 2OH- (aq) + H2 (g) ΔH°rxn = ? H+ (aq) + OH- (aq) → H2O (l) ΔH°rxn, 1 = -56 kJ/mol H2O Zn (s) → Zn2+ (aq) ΔH°rxn, 2 = -153.9 kJ 1/2 H2 (g) → H+ (aq) ΔH°rxn, 3 = 0.0 kJ
Given the following reactions: 2S (s) + 3O2 (g) → 2SO3 (g) H = -840.0 kJ S (s) + O2 (g) → SO2 (g) H = -300.0 kJCalculate the enthalpy of the following reaction in kJ 2SO2 (g) + O2 (g) → 2SO3 (g)
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 industrial process for making sulfuric acid has three steps. Using the data given, calculate the enthalpy change for the overall reaction for the process (the equation is given below).2S(s) + 3O2(g) rightarrow 2H2SO4(l) Data:S(s) + O2(g) → SO(g) Δ Hrxn = -296.83 kJ2SO3(g) → O2(g) + 2SO2(g) ΔHrxn = - 198.2 kJSO3(g) + H2O(l) → H2SO4(l) ΔHrxn = - 227.72 kJSelect one:•-849.9 kJ•-722.8 kJ•-1246.3 kJ•-326.4 kJ
Calculate ΔH°rxn for the reaction 2Ni (s) + 2S (s) + 3O2 (g) → 2NiSO3 (s) from the following information: (1) NiSO3 (s) → NiO (s) + SO3 (g) ΔH°rxn = 156 kJ (2) S (s) + O2 (g) → SO2 (g) ΔH°rxn = - 297 kJ (3) Ni (s) + 1/2O2 (g) → NiO (s) ΔH°rxn = -241 kJ
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.
Enter your answer in the provided box.You are given the following data: H2 (g) → 2H (g) ΔH° = -436.4 kJ/mol Br2 (g) → 2HBr (g) ΔH° = 192.5 kJ/mol H2 (g) + Br2 (g) → 2HBr (g) ΔH° = -72.4 kJ/mol Calculate ΔH° for the reaction H (g) + Br (g) → HBr (g)
Given the following data (1) 2ClF (g) + O2 (g) → Cl2O (g) + OF2 (g) ΔH°rxn = 167.5 kJ (2) 2F2 (g) + O2 (g) → 2OF2 (g) ΔH°rxn = -43.5kJ (3) 2ClF3 (l) + 2O2 (g) → Cl2O (g) + 3OF2 (g) ΔH°rxn = 394.1 kJ Calculate ΔHrxn for the reaction. Show your work! ClF (g) + F2 (g) → ClF3 (l)
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.
Consider the following hypothetical reactions:A → B ΔH = +20kJ B → C ΔH = +66kJ Use Hess's law to calculate the enthalpy change for the reaction A → C. Express your answer using two significant figures.
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 N2 (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