Chemical Energy & Thermodynamics — Chapter Practice, Grade 11

Use calorimetry, Hess’s law, entropy, and free energy with explicit systems and signs.

Problem 1

In an endothermic reaction, the sign of ΔH is…
  1. negative
  2. positive
  3. always zero
  4. impossible to determine
Answer and explanation

positive

An endothermic reaction absorbs energy into the system, so ΔH is positive.

Problem 2

How much heat (in J) is absorbed when 50.0 g of water is warmed from 15.0°C to 35.0°C? (c = 4.18 J/g°C)
Answer and explanation

4180 J

ΔT = 35.0 − 15.0 = 20.0°C, so q = mcΔT = 50 × 4.18 × 20 = 4180 J.

Problem 3

On a potential-energy diagram, the activation energy is the difference in energy between…
  1. the two dashed helper lines
  2. the reactants and the products
  3. the reactants and the peak (the activated complex)
  4. the products and the x-axis
Answer and explanation

the reactants and the peak (the activated complex)

Activation energy is the height of the peak above the reactant energy level — the energy barrier the reaction must climb.

Problem 4

In an insulated, constant-pressure reaction calorimeter, the measured solution temperature falls. Assume negligible heat loss and no other process. Is the reaction endothermic or exothermic, what is the sign of ΔH, and why?
Answer and explanation

Under the stated assumptions, the reaction absorbs heat from the solution, so it is endothermic and ΔH is positive. A temperature fall without these conditions would not uniquely prove reaction enthalpy; other energy transfers or processes could contribute.

Problem 5

A reaction releases 6270 J of heat into 100. g of water (c = 4.18 J/g°C). By how many °C does the water's temperature rise?
Answer and explanation

15 °C

Rearranging q = mcΔT gives ΔT = q ÷ (m × c) = 6270 ÷ (100 × 4.18) = 6270 ÷ 418 = 15°C.

Problem 6

If ΔH=−50 kJ for one reaction amount, find ΔH for twice that amount. Show the reasoning, not just the conclusion.
Hints

Do not omit the reaction-amount basis when scaling.

Answer and explanation

−100 kJ under identical conditions.

Problem 7

Why does bond-energy estimation subtract energy released in forming product bonds from energy used breaking reactant bonds?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Net reaction energy balances input to break bonds against release from forming new bonds; breaking alone is not the energy source.

Problem 8

Reaction enthalpy belongs to the equation’s stated amount and directionReaction enthalpy belongs to the equation’s stated amount and directionA → B ΔH = −40 kJ per equation amount2A → 2B ΔH = −80 kJB → A ΔH = +40 kJScale coefficients and energy together; reversal changes the sign.
Original hypothetical transformation values, not real thermochemical data or an assertion about rate.
Interpret the supplied visual. What is wrong with reversing A→B while keeping the same negative ΔH sign?
Answer and explanation

It reverses the physical energy change without reversing the account’s sign; the corresponding B→A enthalpy is +40 kJ for the displayed basis.

Enthalpy is a state function, and a thermochemical reaction change is associated with a defined transformation and amount basis. Multiplying every coefficient scales the enthalpy change by the same factor; reversing the reaction reverses its sign. A number labeled kJ/mol must specify mol of what—product, reactant or reaction extent. Phase labels also matter because liquid and vapor have different enthalpy states. Keep these details when using a reference value in stoichiometry.

It reverses the physical energy change without reversing the account’s sign; the corresponding B→A enthalpy is +40 kJ for the displayed basis.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 9

A selected reaction absorbs 720 kJ in the bond-breaking account and releases 880 kJ on product formation. Estimate net enthalpy, then predict the corresponding value when the entire equation is doubled.
Answer and explanation

ΔH≈720−880=−160 kJ for the selected equation amount. Doubling all coefficients gives approximately −320 kJ. Average bond energies are an approximate full-reaction estimate, not proof about spontaneity or rate.

ΔH≈720−880=−160 kJ for the selected equation amount. Doubling all coefficients gives approximately −320 kJ. Average bond energies are an approximate full-reaction estimate, not proof about spontaneity or rate.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 10

OpenSciEd magnification template linking the observable heater to two particle-scale views before and after adding water; it is a model-building prompt, not a completed reaction diagram.
OpenSciEd’s empty observable/particle-scale heater template is a model-building prompt, not a completed reaction or a calorimeter record. Flameless-heater model at visible and particle scales · OpenSciEd · 7.2 Chemical Reactions & Energy, Lesson 1, source, CC BY 4.0. Figure excerpt rendered from the PDF and resized for the web.
What would a complete explanation need to add to the empty source heater template? Distinguish system boundary, reaction species, bond-energy account and thermometer response, and name one conclusion the template alone cannot establish.
Answer and explanation

Define the selected reaction system and surroundings, identify species from appropriate evidence, conserve material and account for bond breaking plus formation. Connect transferred energy with heat capacity and measured temperature under stated conditions. The empty template alone cannot identify the reaction, quantify enthalpy or prove a construction method.

Identify the source’s visible labels, scales and unfinished or omitted information.

Use the local complete explanations to connect the observation with an explicit mechanism.

State assumptions and a limit before extending the model.

Response criteria

  • Interpret the source symbols and boundary correctly.
  • Explain the relevant mechanism using quantities/species consistently.
  • State a meaningful omission, assumption or evidence limitation.

Problem 11

Reaction warms 50 g water by 4°C; c=4.18. Find heat gained by water.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

836 J

50×4.18×4.

Problem 12

Why reverse ΔH when reversing a reaction?
Answer and explanation

The same state difference is traversed in the opposite direction.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 13

For 40 g water warming 5°C with c=4.18, find water heat gain.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

836 J

40×4.18×5=836.

Problem 14

How does neglected heat loss affect an exothermic cup estimate?
Answer and explanation

Less heat reaches the measured solution, so the estimated release magnitude is typically too small.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 15

A reaction loses 3.0 kJ to water and 0.5 kJ to its container. Find its total heat change. Show the reasoning, not just the conclusion.
Hints

System and surroundings have opposite signs in this boundary.

Answer and explanation

−3.5 kJ, assuming no other transfers.

Problem 16

Two steps have ΔH=+20 and −55 kJ and add to a target reaction. Find target ΔH.
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

20−55=−35 kJ by Hess's law with the given orientations and scales.

Problem 17

Hess’s law combines state changes while canceling intermediatesHess’s law combines state changes while canceling intermediatesA → B ΔH₁ = −30 kJB → C ΔH₂ = +12 kJA → C ΔH = −18 kJCancel intermediates and add the corresponding energy changes.
Original hypothetical two-step enthalpy cycle with stated values and compatible species states assumed.
Interpret the supplied visual. Why can B cancel while the +12 kJ second step must still be included?
Answer and explanation

B is an intermediate in the summed transformation. Its associated state-change energy remains part of the overall endpoint balance.

Hess’s law uses the state-function character of enthalpy: the overall change depends on initial and final states rather than which conceptual route is added. Scale or reverse component equations as necessary, cancel shared intermediates and apply the same operations to their enthalpies. Intermediate B can disappear from the net equation without implying it was never involved in a real mechanism. The algebraic route need not be the actual laboratory pathway, and phases must match when canceling a species.

B is an intermediate in the summed transformation. Its associated state-change energy remains part of the overall endpoint balance.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 18

A reaction heats 50.0 g water by 8.00°C. Use c=4.18 J/(g·°C), neglect apparatus/losses, and assume 0.0200 mol reacts. Estimate reaction heat and molar enthalpy on that basis.
Answer and explanation

qwater=50.0×4.18×8.00=1672 J. qreaction≈−1.672 kJ. Dividing by 0.0200 mol gives approximately −83.6 kJ/mol on the specified reacting-amount basis, subject to the neglected terms.

qwater=50.0×4.18×8.00=1672 J. qreaction≈−1.672 kJ. Dividing by 0.0200 mol gives approximately −83.6 kJ/mol on the specified reacting-amount basis, subject to the neglected terms.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 19

ΔH=−50kJ/mol, ΔS=−.100kJ/mol·K, T=300 K. Find ΔG.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

-20 kJ/mol

−50−300(−.100)=−20.

Problem 20

Can a favorable reaction be slow?
Answer and explanation

Yes. A high activation-energy barrier can limit its rate.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 21

ΔH=+12 kJ/mol, ΔS=+.050 kJ/(mol·K), T=300 K. Find ΔG.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

-3 kJ/mol

12−300(.050)=−3.

Problem 22

ΔH=30 kJ/mol, ΔS=.050 kJ/mol·K. Find crossover T.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

600 K

30/.050=600.

Problem 23

For ΔH=−20 kJ/mol and ΔS=−0.040 kJ/(mol·K) at 300 K, find ΔG. Show the reasoning, not just the conclusion.
Hints

Units and absolute temperature must match.

Answer and explanation

−20−300×(−0.040)=−8 kJ/mol.

Problem 24

Why can an endothermic process be thermodynamically favorable at high temperature?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

A sufficiently positive entropy change makes TΔS exceed ΔH, producing negative ΔG under appropriate conditions.

Problem 25

Standard-state favorability and the actual mixture are differentStandard-state favorability and the actual mixture are differentΔG = ΔG° + RT ln QAt equilibrium: ΔG = 0 and Q = KΔG° = −RT ln KQ and K use compatible dimensionless activity conventions.
Original thermodynamic relationships; no logarithm of a dimensional raw concentration is intended.
Interpret the supplied visual. Why can a reaction with negative ΔG° fail to proceed net forward from a very product-rich starting mixture?
Answer and explanation

A sufficiently large Q can make RT ln Q positive enough to change the actual ΔG; the mixture composition matters.

The standard reaction free energy ΔG° refers to a defined reference-state account. The actual driving tendency also depends on mixture composition through Q, the reaction quotient, using consistent dimensionless activities. At equilibrium Q=K and the reaction’s ΔG is zero. A negative ΔG° does not mean every possible starting mixture moves forward, because the RT ln Q contribution can change the actual sign. Concentration-only classroom expressions approximate activities and require an explicitly consistent convention.

A sufficiently large Q can make RT ln Q positive enough to change the actual ΔG; the mixture composition matters.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 26

For a selected process use ΔH=+30.0 kJ/mol and ΔS=+0.150 kJ/(mol·K), treated constant. Find ΔG at 250 K, the modeled crossover temperature and one limit of the prediction.
Answer and explanation

At 250 K, ΔG=30.0−250(0.150)=−7.50 kJ/mol. The modeled crossover is 30.0/0.150=200 K. Actual ΔH/ΔS can vary with temperature and phases, and the calculation does not establish a rate.

At 250 K, ΔG=30.0−250(0.150)=−7.50 kJ/mol. The modeled crossover is 30.0/0.150=200 K. Actual ΔH/ΔS can vary with temperature and phases, and the calculation does not establish a rate.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Browse all courses