Energy Cycles In Reactions

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Bond Enthalpy Basics

A chemical bond is a force of attraction between two atoms.
Breaking a bond requires energy input, so it is an endothermic process.
Forming a bond releases energy, so it is an exothermic process.
The energy required to break a particular bond is called the bond dissociation enthalpy, often shortened to bond enthalpy or bond energy.
The energy released when a bond forms has the same magnitude as the energy taken in to break it, but with opposite sign.

Overall Enthalpy Changes

If more energy is released when new bonds form than is required to break bonds, the reaction is exothermic.
In an exothermic reaction, the products are more stable than the reactants.
If more energy is required to break bonds than is released when new bonds form, the reaction is endothermic.
In an endothermic reaction, the products are less stable than the reactants.

Exothermic and endothermic reactions

Average Bond Enthalpy

Bond energies are affected by other atoms in the molecule (the environment).
An average bond enthalpy is defined as the energy needed to break one mole of bonds in a gaseous molecule, averaged over similar compounds.
For example, the average C–H bond enthalpy is found by taking the bond dissociation enthalpy for the whole molecule and dividing by the number of C–H bonds.
The first C–H bond is easier to break than the second because the remaining hydrogens are pulled more closely to the carbon.
Since it is impossible to measure the energy of each individual C–H bond, an average is taken and compared with similar compounds to obtain an accepted value.

Bond Enthalpy Calculations

Bond energies are used to find the ΔH of a reaction when it cannot be done experimentally.
The process is a step-by-step summation of the bond enthalpies of all molecules present.
The formula is: ΔH = enthalpy change for bonds broken + enthalpy change for bonds formed.
Values for bonds broken are positive (endothermic) and values for bonds formed are negative (exothermic).
Always draw out full displayed structures to account for every bond present.
Watch out for coefficients in balanced equations; multiply bond enthalpies by the coefficients.
Show your steps because bond enthalpy calculations often carry marks for workings.

Hess's Law

In 1840, Russian chemist Germain Hess formulated Hess's Law.
Hess's Law states: 'The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same.'
This means whether a reaction takes place in one or two steps, the total enthalpy change is the same.
Hess's Law is used to calculate enthalpy changes that cannot be found experimentally using calorimetry.
For example, ΔHf of propane cannot be found experimentally as hydrogen and carbon don't react under standard conditions.
Hess's Law is based on the Law of Conservation of Energy: energy cannot be created or destroyed, only changed in form.

Calculating ΔH from ΔHf using Hess's Law Energy Cycles

According to Hess's Law, the total enthalpy change is the same regardless of the pathway taken, because energy is conserved.
Two possible routes to form products from elements: direct formation (one-step) and indirect formation (two-step via reactants).
Direct route: Elements → Products, enthalpy change = ΔH₂.
Indirect route: Elements → Reactants → Products, enthalpy change = ΔH₁ + ΔHr.
By Hess's Law: ΔH₂ = ΔH₁ + ΔHr.
Rearranged to solve for enthalpy change of reaction: ΔHr = ΔH₂ − ΔH₁.

Hess's Law Calculations Using Cycles

There are two common methods to solve Hess's Law problems: using cycles and using equations.
To use cycles, follow a step-by-step plan: write the target enthalpy change at the top, write the alternative route at the bottom, connect with arrows, and add enthalpy data adjusting for molar amounts.
Two important rules: if you follow the direction of the arrow, ADD the quantity; if you go against the arrow, SUBTRACT the quantity.
Example: Calculate ΔH for 2N₂(g) + 6H₂(g) → 4NH₃(g) given 4NH₃(g) + 3O₂(g) → 2N₂(g) + 6H₂O(l) ΔH = -1530 kJ mol⁻¹ and H₂(g) + ½O₂(g) → H₂O(l) ΔH = -288 kJ mol⁻¹.
Solution: ΔH = +6ΔH₂ – ΔH₁ = +(-288 × 6) - (-1530) = -198 kJ.

Hess's Law Calculations Using Equations

When using equations, identify which given equation contains the desired product and reactant.
Adjust equations if necessary: if you reverse an equation, reverse the sign of ΔH; if you multiply an equation, multiply ΔH by the same factor.
Add the equations together and cancel common items on both sides.
Add the ΔH values together to get the final enthalpy change.
Example: Given N₂(g) + O₂(g) → 2NO(g) ΔH = +180 kJ and 2NO(g) → 2NO₂(g) + O₂(g) ΔH = +112 kJ, find ΔH for N₂(g) + 2O₂(g) → 2NO₂(g).
Reverse the second equation: 2NO₂(g) + O₂(g) → 2NO(g) ΔH = -112 kJ. Add to first: N₂(g) + O₂(g) + 2NO₂(g) + O₂(g) → 2NO(g) + 2NO₂(g). Cancel: N₂(g) + 2O₂(g) → 2NO₂(g) ΔH = +180 - 112 = +68 kJ.

Examiner Tips and Tricks

You do not need to learn Hess's Law word for word, but you must understand the principle as it provides the foundation for problem solving in Chemical Energetics.
When using cycles, always put brackets around values and add the mathematical operator in front to avoid confusing signs with operations.
It doesn't matter whether you use equations or cycles to solve Hess's Law problems, but you should be familiar with both methods and sometimes one is easier than another.

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연습 문제

무료 미리 보기 — 61개 중 8개 문제. 가입하면 전부 볼 수 있어요.

1.Which statement best describes what is meant by the average HI bond enthalpy?
Easy
- AThe energy stored in a covalent bond.
- BThe energy required to break one covalent bond in the gas phase.
- CThe energy required to break one mole of the HI bonds in the gas phase.
- DThe energy released when two atoms form a covalent bond.
2.Which is the correct definition of mean bond enthalpy?
Easy
- AThe amount of energy required to break a specific covalent bond in the gas phase
- BThe energy required to break one mole of a specific covalent bond with all chemicals in their standard states
- CThe amount of energy required to break a specific covalent bond with all chemicals in their standard states
- DThe energy required to break one mole of a specific covalent bond in the gas phase
3.A basic definition of Hess’s Law states that the overall enthalpy change for a reaction is the same independent of the route taken. Which of the following statements make the definition of Hess’s Law more complete? (select all that apply)
Medium
- AProviding that the reactants are the same
- BProviding that the products are the same
- CProviding that the conditions at the start and the end of the reaction are the same
- DProviding that the reaction is exothermic
- EProviding that the reaction goes to completion
4.Which equation shows the correct application of Hess’s law to calculate the enthalpy change for the conversion of graphite to diamond, given the enthalpy changes of formation of graphite (ΔH₁) and diamond (ΔH₂) from their elements?
Medium
- AΔH = ΔH₁ + ΔH₂
- BΔH = ΔH₁ - ΔH₂
- CΔH = ΔH₂ - ΔH₁
- DΔH = ΔH₁ × ΔH₂
5.A student drew a Hess cycle to calculate the enthalpy of reaction to produce ethane from ethene and hydrogen. The student used the following enthalpy of combustion data: C₂H₄(g) ΔHc = -1411 kJ mol⁻¹, H₂(g) ΔHc = -286 kJ mol⁻¹, C₂H₆(g) ΔHc = -1560 kJ mol⁻¹. What are the correct labels for the arrows for the student’s Hess cycle?
Medium
- AArrow 1: ΔHr, Arrow 2: -1411, Arrow 3: -286, Arrow 4: -1560
- BArrow 1: ΔHr, Arrow 2: -1411, Arrow 3: +286, Arrow 4: -1560
- CArrow 1: ΔHr, Arrow 2: +1411, Arrow 3: -286, Arrow 4: +1560
- DArrow 1: ΔHr, Arrow 2: -1411, Arrow 3: -286, Arrow 4: +1560
6.The diagram shows two possible reaction pathways for the reaction of A → D. Which of the following statements are correct? I. A → D ΔH = +45 kJ II. C → D ΔH = -25 kJ III. D → C ΔH = -65 kJ
Medium
- AI and II only
- BI and III only
- CII and III only
- DI, II and III
7.The thermal decomposition of calcium carbonate is very slow and requires a high temperature. The enthalpy change can be determined by two reactions with dilute hydrochloric acid. Which set of chemicals correctly completes the Hess cycle diagram?
Medium
- ACaCO₃ + CaO + HCl
- BCaCl₂ + H₂O + CO₂
- CCaCl₂ + H₂O
- DCaCl₂ + H₂O + CO₂ + HCl
8.The hydration enthalpy of anhydrous copper(II) sulfate cannot be measured directly. It can be found indirectly by determining the solution enthalpies of anhydrous and hydrated copper(II) sulfate. Which are the correct labels for the enthalpies shown in the Hess cycle?
Medium
- A1: ΔHsol, 2: ΔHhyd, 3: ΔHhyd
- B1: ΔHr, 2: ΔHhyd, 3: ΔHhyd
- C1: ΔHhyd, 2: ΔHsol, 3: ΔHhyd
- D1: ΔHr, 2: ΔHsol, 3: ΔHsol

더 많은 문제

9.The hydration enthalpy of anhydrous copper(II) sulfate, ΔHexp, cannot be measured directly. Which of the following statements correctly explain why? (select all that apply)
Medium
- AMeasuring the temperature change in a solid is difficult
- BThe reaction is very slow
- CThe reaction has high energy requirements
- DThe reaction is endothermic
- EThe reaction does not occur under standard conditions
10.The incomplete combustion of methanol can produce carbon monoxide and water: CH₃OH + O₂ → CO + 2H₂O. Using the mean bond dissociation enthalpies: C-H 414, C-O 358, O-H 463, O=O 498, C≡O 1072 kJ mol⁻¹, what is the enthalpy change for this incomplete combustion of methanol?
Medium
- A-363 kJ mol⁻¹
- B-466 kJ mol⁻¹
- C-288 kJ mol⁻¹
- D-144 kJ mol⁻¹
11.Some bond energy values are: Br–Br 193, Cl–Cl 242, C–H 414, C–Cl 324 kJ mol⁻¹. These relate to reactions: W: Br₂ → 2Br, X: 2Cl → Cl₂, Y: CH₄ + Cl₂ → CH₃Cl + HCl, Z: CH₄ → CH₃ + H. What is the correct order of enthalpy changes from most negative to most positive?
Hard
- Y
- Z
- W
- X
12.Shown below are three enthalpy changes: CH₄(g) + O₂(g) → HCHO(l) + H₂O(l) ΔH = x; HCHO(l) + ½O₂(g) → HCOOH(l) ΔH = y; 2HCOOH(l) + ½O₂(g) → (COOH)₂(l) + H₂O(l) ΔH = z. What is the correct expression for the enthalpy change of: 2CH₄(g) + 3½O₂(g) → (COOH)₂(l) + 3H₂O(l)?
Hard
- Ax + y + z
- B2x + y + z
- C2x + 2y + z
- D2x + 2y + 2z
13.Breaking a chemical bond is an endothermic process.
Easy
True or false?
14.The energy released when a bond is formed is always greater than the energy required to break the same bond.
Easy
True or false?
15.Match each term with its correct description.
Medium
- Bond breaking
- Bond formation
- Hess's Law
- Endothermic process
- Exothermic process
- Enthalpy change is independent of route
16.Which statement about bond enthalpy is correct?
Medium
- AThe energy required to break a bond is always greater than the energy released when the same bond forms.
- BThe energy released when a bond forms is equal in magnitude but opposite in sign to the energy required to break it.
- CBond enthalpy is always positive for bond formation.
- DBond enthalpy depends only on the atoms involved and not on the molecular environment.
17.In a Hess's Law cycle, if you follow the direction of an arrow, what operation should you apply to the enthalpy change value?
Medium
- AAdd the quantity
- BSubtract the quantity
- CMultiply the quantity
- DReverse the sign of the quantity
18.Which of the following statements about bond enthalpy calculations are correct? (select all that apply)
Medium
- AThe enthalpy change for bonds broken is positive.
- BThe enthalpy change for bonds formed is negative.
- CΔHr = bonds broken + bonds formed.
- DThe coefficients in the balanced equation must be multiplied by the bond enthalpies.
- EAverage bond enthalpies are always exact for a specific molecule.
19.Hess's Law is a consequence of the Law of Conservation of Energy.
Easy
True or false?
20.Which of the following is the correct expression for the enthalpy change of a reaction using bond enthalpies?
Medium
- AΔH = Σ(bonds broken) - Σ(bonds formed)
- BΔH = Σ(bonds formed) - Σ(bonds broken)
- CΔH = Σ(bonds broken) + Σ(bonds formed)
- DΔH = Σ(bonds formed) / Σ(bonds broken)
21.Which of the following statements about bond enthalpy are correct? (select all that apply)
Medium
- ABreaking a bond is endothermic.
- BForming a bond is exothermic.
- CThe energy released when a bond is formed is always greater than the energy required to break it.
- DThe energy required to break a bond depends on the atoms forming the bond.
- EAverage bond enthalpies are always exact for a specific bond in a molecule.
22.Arrange the following steps in the correct order to calculate the enthalpy change of a reaction using bond enthalpies.
Medium
- Write the balanced chemical equation.
- Draw the full displayed structures of all molecules.
- Sum the bond enthalpies of bonds broken.
- Sum the bond enthalpies of bonds formed.
- Calculate ΔH = (bonds broken) + (bonds formed).
23.Which of the following statements about Hess's Law are correct? (select all that apply)
Medium
- AThe total enthalpy change is independent of the route taken.
- BThe initial and final conditions must be the same.
- CHess's Law is a consequence of the Law of Conservation of Energy.
- DHess's Law can only be used for reactions that occur quickly.
- EHess's Law can be used to calculate enthalpy changes that cannot be measured directly.
24.Match each enthalpy change with its correct definition.
Medium
- Standard enthalpy of formation
- Standard enthalpy of combustion
- Standard enthalpy of solution
- Standard enthalpy of hydration
- Enthalpy change when one mole of a compound is formed from its elements in their standard states.
- Enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions.
- Enthalpy change when one mole of a solute dissolves in a solvent to form an infinitely dilute solution.
- Enthalpy change when one mole of gaseous ions is hydrated to form aqueous ions.
25.Arrange the following steps in the correct order to solve a Hess's Law problem using equations.
Medium
- Identify the target equation.
- Find the given equations that contain the reactants and products.
- Reverse equations if necessary and change the sign of ΔH.
- Multiply equations to match the number of moles and multiply ΔH.
- Add the equations and cancel common terms.
- Add the ΔH values to get the final enthalpy change.
26.Which of the following are valid reasons for using Hess's Law? (select all that apply)
Medium
- ATo calculate enthalpy changes that cannot be measured directly by calorimetry.
- BTo determine the enthalpy change of formation of propane, which cannot be found experimentally under standard conditions.
- CTo find the enthalpy change of a reaction that is too slow to measure.
- DTo calculate the enthalpy change of a reaction that does not occur under standard conditions.
- ETo measure the exact bond enthalpy of a specific C-H bond in methane.
27.Match each quantity with its correct expression in a Hess's Law cycle.
Medium
- ΔH for direct route
- ΔH for indirect route via elements
- Enthalpy change of reaction from formation enthalpies
- Enthalpy change of reaction from combustion enthalpies
- ΔH = ΔHf(products) - ΔHf(reactants)
- ΔH = ΔHc(reactants) - ΔHc(products)
- ΔH = ΔH1 + ΔH2
- ΔH = ΔH2 - ΔH1
28.Arrange the following steps in the correct order to solve a Hess's Law problem using a cycle.
Hard
- Write the target enthalpy change at the top of the diagram.
- Write the alternative route at the bottom of the cycle.
- Add the enthalpy data and adjust for different molar amounts.
- Write the Hess's Law calculation using the rules: follow arrow = add, against arrow = subtract.
29.Which of the following statements about average bond enthalpies are correct? (select all that apply)
Hard
- AThey are averaged over similar compounds because bond energy depends on the molecular environment.
- BThe first C-H bond in methane is easier to break than the second.
- CThey can be used to calculate enthalpy changes for reactions that cannot be measured directly.
- DThey are exact for each individual bond in a molecule.
- EThey are always positive values.
30.Arrange the following reactions in order of increasing enthalpy change (most negative to most positive) based on bond energies: Br–Br 193, Cl–Cl 242, C–H 414, C–Cl 324 kJ mol⁻¹. Reactions: W: Br₂ → 2Br, X: 2Cl → Cl₂, Y: CH₄ + Cl₂ → CH₃Cl + HCl, Z: CH₄ → CH₃ + H.
Hard
- W
- X
- Y
- Z
31.Which of the following are correct statements about the relationship between bond breaking and bond making? (select all that apply)
Medium
- AIf more energy is released when new bonds are formed than is required to break bonds, the reaction is exothermic.
- BIf more energy is required to break bonds than is released when new bonds are formed, the reaction is endothermic.
- CThe products are more stable than the reactants in an exothermic reaction.
- DThe products are more stable than the reactants in an endothermic reaction.
- EThe energy released when a bond is formed is always greater than the energy required to break the same bond.
32.The equation for the complete combustion of ethyne is 2C₂H₂(g) + 5O₂(g) → 2H₂O(g) + 4CO₂(g). Using the average bond enthalpies C−H 414, C≡C 839, O=O 498, C=O 804 and O−H 463 kJ mol⁻¹, what is the enthalpy of combustion of ethyne in kJ mol⁻¹?
Medium
- A-1230
- B-615
- C-2460
- D+1230
33.The equation for the incomplete combustion of methanol is CH₃OH + O₂ → CO + 2H₂O. Using the mean bond dissociation enthalpies C-H 414, C-O 358, O-H 463, O=O 498 and C≡O 1072 kJ mol⁻¹, what is the enthalpy change for this reaction in kJ mol⁻¹?
Medium
- A-466
- B-288
- C-144
- D-368
34.The standard enthalpy change for N₂(g) + 3F₂(g) → 2NF₃(g) is -246 kJ. Given N≡N = 945 kJ mol⁻¹ and F-F = 159 kJ mol⁻¹, what is the bond energy of the N-F bond in kJ mol⁻¹?
Medium
- A196
- B588
- C278
- D1668
35.Given I₂(s) → I₂(g) ΔH = +38 kJ mol⁻¹ and I₂(g) + 3Cl₂(g) → 2ICl₃(s) ΔH = -214 kJ mol⁻¹, what is the standard enthalpy of formation of ICl₃(s) in kJ mol⁻¹?
Medium
- A-88
- B-352
- C352
- D88
36.Using CO(g) + ½O₂(g) → CO₂(g) ΔH = -283 kJ mol⁻¹; H₂(g) + ½O₂(g) → H₂O(l) ΔH = -286 kJ mol⁻¹; H₂O(g) → H₂O(l) ΔH = -44 kJ mol⁻¹, what is the enthalpy change for CO(g) + H₂(g) → CO₂(g) + H₂O(g) in kJ mol⁻¹?
Medium
- A-525
- B-613
- C41
- D-47
37.Iodine trichloride is made by: I₂(s) + Cl₂(g) → 2ICl(s) ΔH = +14 kJ mol⁻¹; ICl(s) + Cl₂(g) → ICl₃(s) ΔH = -88 kJ mol⁻¹. What is the enthalpy change of formation of solid iodine trichloride in kJ mol⁻¹?
Medium
- A-81
- B-74
- C-162
- D-60
38.Given: CH₄(g) + O₂(g) → HCHO(l) + H₂O(l) ΔH = x; HCHO(l) + ½O₂(g) → HCOOH(l) ΔH = y; 2HCOOH(l) + ½O₂(g) → (COOH)₂(l) + H₂O(l) ΔH = z. What is the enthalpy change for 2CH₄(g) + 3½O₂(g) → (COOH)₂(l) + 3H₂O(l)?
Hard
- A2x + 2y + z
- Bx + y + z
- C2x + y + z
- D2x + 2y + 2z
39.The diagram shows two possible reaction pathways for A → D. Path 1: A → B ΔH = +20 kJ, B → C ΔH = -30 kJ, C → D ΔH = +55 kJ. Path 2: A → D directly. Which of the following statements are correct? I. A → D ΔH = +45 kJ; II. C → D ΔH = -25 kJ; III. D → C ΔH = -65 kJ.
Hard
- AI and III only
- BI and II only
- CII and III only
- DI, II and III
40.Some bond energy values are: Br–Br 193, Cl–Cl 242, C–H 414, C–Cl 324 kJ mol⁻¹. These relate to: W: Br₂ → 2Br; X: 2Cl → Cl₂; Y: CH₄ + Cl₂ → CH₃Cl + HCl; Z: CH₄ → CH₃ + H. What is the correct order of enthalpy changes from most negative to most positive?
Hard
- AX → Y → Z → W
- BY → Z → W → X
- CZ → W → X → Y
- DY → X → W → Z
41.The standard enthalpy change for the reaction N₂(g) + 3F₂(g) → 2NF₃(g) is -246 kJ. The bond energy of N≡N is 945 kJ mol⁻¹ and F–F is 159 kJ mol⁻¹. What is the bond energy of the N–F bond in kJ mol⁻¹?
Medium
42.Given: I₂(s) → I₂(g) ΔH = +38 kJ mol⁻¹; I₂(g) + 3Cl₂(g) → 2ICl₃(s) ΔH = -214 kJ mol⁻¹. What is the standard enthalpy of formation of ICl₃(s) in kJ mol⁻¹?
Hard
43.Using: CO(g) + ½O₂(g) → CO₂(g) ΔH = -283 kJ mol⁻¹; H₂(g) + ½O₂(g) → H₂O(l) ΔH = -286 kJ mol⁻¹; H₂O(g) → H₂O(l) ΔH = -44 kJ mol⁻¹. What is the enthalpy change for CO(g) + H₂O(g) → CO₂(g) + H₂(g) in kJ mol⁻¹?
Hard
44.Iodine trichloride, ICl₃, is made by: I₂(s) + Cl₂(g) → 2ICl(s) ΔH = +14 kJ mol⁻¹; ICl(s) + Cl₂(g) → ICl₃(s) ΔH = -88 kJ mol⁻¹. What is the enthalpy change of formation for solid iodine trichloride in kJ mol⁻¹?
Hard
45.Complete the definition of average bond energy: 'The energy needed to break one ___ of bonds in a ___ molecule averaged over similar compounds.'
Easy
The energy needed to break one ____ of bonds in a ____ molecule averaged over similar compounds.
46.What term describes the energy required to break a particular bond in a gaseous molecule?
Easy
47.Complete the statement of Hess's Law: 'The total enthalpy change in a chemical reaction is ___ of the route by which the chemical reaction takes place as long as the ___ and final conditions are the same.'
Medium
The total enthalpy change in a chemical reaction is ____ of the route by which the chemical reaction takes place as long as the ____ and final conditions are the same.
48.Using the bond energies: N≡N 945, H−H 436, N−H 391 kJ mol⁻¹, calculate the enthalpy change for the Haber process: N₂(g) + 3H₂(g) → 2NH₃(g). Give your answer in kJ mol⁻¹.
Medium
49.Using the bond energies N≡N 945, H−H 436, N−H 391 kJ mol⁻¹, calculate the enthalpy change for the Haber process N₂(g) + 3H₂(g) → 2NH₃(g) in kJ mol⁻¹.
Hard
50.Calculate the enthalpy change, in kJ mol⁻¹, for the reaction 4FeO(s) + O₂(g) → 2Fe₂O₃(s) given: 2Fe(s) + O₂(g) → 2FeO(s) ΔH = -544 kJ; 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1648 kJ.
Hard
51.Complete the statement of Hess's Law.
Medium
The total enthalpy change in a chemical reaction is ____ of the route by which the chemical reaction takes place as long as the initial and final ____ are the same.
52.Complete the definition of average bond energy.
Medium
The energy needed to break one ____ of bonds in a ____ molecule averaged over similar compounds.
53.Explain why the enthalpy change for the reaction 3C(s) + 4H₂(g) → C₃H₈(g) cannot be found experimentally under standard conditions.
Medium
54.Explain why average bond enthalpies are used instead of exact bond dissociation enthalpies for a specific bond in a molecule like methane.
Hard
55.The standard enthalpy change for the reaction N₂(g) + 3F₂(g) → 2NF₃(g) is -246 kJ. The bond energy of N≡N is 945 kJ mol⁻¹ and F–F is 159 kJ mol⁻¹. What is the bond energy of the N–F bond in kJ mol⁻¹?
Medium
- A306 kJ mol⁻¹
- B278 kJ mol⁻¹
- C250 kJ mol⁻¹
- D334 kJ mol⁻¹
56.Given: I₂(s) → I₂(g) ΔH = +38 kJ mol⁻¹; I₂(g) + 3Cl₂(g) → 2ICl₃(s) ΔH = -214 kJ mol⁻¹. What is the standard enthalpy of formation of ICl₃(s) in kJ mol⁻¹?
Hard
- A-79 kJ mol⁻¹
- B-88 kJ mol⁻¹
- C-70 kJ mol⁻¹
- D-97 kJ mol⁻¹
57.Using: CO(g) + ½O₂(g) → CO₂(g) ΔH = -283 kJ mol⁻¹; H₂(g) + ½O₂(g) → H₂O(l) ΔH = -286 kJ mol⁻¹; H₂O(g) → H₂O(l) ΔH = -44 kJ mol⁻¹. What is the enthalpy change for CO(g) + H₂O(g) → CO₂(g) + H₂(g) in kJ mol⁻¹?
Hard
- A-41 kJ mol⁻¹
- B-33 kJ mol⁻¹
- C-45 kJ mol⁻¹
- D-37 kJ mol⁻¹
58.Iodine trichloride, ICl₃, is made by: I₂(s) + Cl₂(g) → 2ICl(s) ΔH = +14 kJ mol⁻¹; ICl(s) + Cl₂(g) → ICl₃(s) ΔH = -88 kJ mol⁻¹. What is the enthalpy change of formation for solid iodine trichloride in kJ mol⁻¹?
Hard
- A-65 kJ mol⁻¹
- B-81 kJ mol⁻¹
- C-89 kJ mol⁻¹
- D-73 kJ mol⁻¹
59.Using the bond energies: N≡N 945, H−H 436, N−H 391 kJ mol⁻¹, calculate the enthalpy change for the Haber process: N₂(g) + 3H₂(g) → 2NH₃(g). Give your answer in kJ mol⁻¹.
Medium
- A-93 kJ mol⁻¹
- B-84 kJ mol⁻¹
- C-102 kJ mol⁻¹
- D-75 kJ mol⁻¹
60.Using the bond energies N≡N 945, H−H 436, N−H 391 kJ mol⁻¹, calculate the enthalpy change for the Haber process N₂(g) + 3H₂(g) → 2NH₃(g) in kJ mol⁻¹.
Hard
- A-93 kJ mol⁻¹
- B-84 kJ mol⁻¹
- C-102 kJ mol⁻¹
- D-75 kJ mol⁻¹
61.Calculate the enthalpy change, in kJ mol⁻¹, for the reaction 4FeO(s) + O₂(g) → 2Fe₂O₃(s) given: 2Fe(s) + O₂(g) → 2FeO(s) ΔH = -544 kJ; 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1648 kJ.
Hard
- A-504 kJ mol⁻¹
- B-560 kJ mol⁻¹
- C-616 kJ mol⁻¹
- D-448 kJ mol⁻¹

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