RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Chemistry · 10 questions · 24 marks
Thermodynamics answers two questions a chemist always cares about: how much energy does a reaction give out, and will it go at all? You will learn to track heat and work through the first law, add up enthalpies by Hess's law, and finally combine enthalpy with entropy in the Gibbs energy equation that decides spontaneity.
For the reaction N₂(g) + 3H₂(g) → 2NH₃(g) carried out at temperature T, the relation between ΔH and ΔU is:
Answer
ΔH = ΔU − 2RT. The relation is ΔH = ΔU + Δn_g RT, where Δn_g counts only gaseous species. Here the products have 2 mol of gas and the reactants have 1 + 3 = 4 mol of gas. So Δn_g = 2 − 4 = −2, and ΔH = ΔU + (−2)RT = ΔU − 2RT. Because the reaction contracts the gas volume, the surroundings do work on the system, which makes the enthalpy change more negative than the internal energy change.
An ideal gas expands into an evacuated container at constant temperature. For this process:
Answer
w = 0, q = 0 and ΔU = 0. Expansion into a vacuum is a free expansion, so the external pressure is zero and w = −p_ext ΔV = 0 no matter how large the volume change is. For an ideal gas the internal energy depends only on temperature, so an isothermal change gives ΔU = 0. Applying the first law, ΔU = q + w gives 0 = q + 0, so q = 0 as well. No heat flows and no work is done, yet the process is strongly spontaneous because the entropy of the gas increases.
Which of the following processes has the largest positive entropy change?
Answer
CaCO₃(s) → CaO(s) + CO₂(g) has the largest positive entropy change. Entropy tracks the freedom available to the particles, and the single biggest factor is the change in the number of moles of gas. Here a solid decomposes to give a solid plus one whole mole of gas, so Δn_g = +1 and a highly ordered lattice releases freely moving gas molecules. The first and fourth options both reduce the moles of gas (3 → 2 and 4 → 2), so their ΔS values are negative. Freezing water converts a mobile liquid into an ordered crystal, which is also a decrease in entropy.
A reaction has ΔH = +58 kJ/mol and ΔS = +160 J K⁻¹ mol⁻¹. This reaction is:
Answer
Spontaneous only above about 363 K. Spontaneity needs ΔG = ΔH − TΔS to be negative. With ΔH positive, the term −TΔS must be large enough in magnitude to overcome it, and since ΔS is also positive, raising T does exactly that. The crossover comes at ΔG = 0, so T = ΔH ÷ ΔS = 58000 J mol⁻¹ ÷ 160 J K⁻¹ mol⁻¹ = 362.5 K, about 363 K. Below this temperature the enthalpy term dominates and the reaction is non-spontaneous; above it the entropy term wins and the reaction proceeds. Note the units of ΔH had to be converted from kJ to J to match ΔS.
Assertion (A): The standard enthalpy of formation of graphite is zero but that of diamond is +1.9 kJ/mol. Reason (R): The standard enthalpy of formation of an element is defined as zero only for its most stable allotrope under standard conditions.
Answer
Both A and R are true and R is the correct explanation of A. Enthalpy has no absolute zero point, so chemists fix an arbitrary but consistent reference: every element in its most stable form at 1 bar and the stated temperature is assigned ΔH°_f = 0. For carbon that most stable form is graphite, so graphite gets the zero. Diamond is a different allotrope and is slightly higher in enthalpy, so converting graphite to diamond is endothermic and diamond carries a small positive standard enthalpy of formation. The reason therefore states exactly the convention that produces the two values in the assertion.
Calculate the work done when 2.0 mol of an ideal gas expands isothermally and reversibly from 5.0 L to 15.0 L at 300 K. (R = 8.314 J K⁻¹ mol⁻¹)
Answer
For an isothermal reversible expansion, w = −2.303 nRT log(V₂ ÷ V₁). Substituting: w = −2.303 × 2.0 × 8.314 × 300 × log(15.0 ÷ 5.0) = −2.303 × 2.0 × 8.314 × 300 × log 3. Since log 3 = 0.4771, w = −2.303 × 2.0 × 8.314 × 300 × 0.4771 = −5.48 × 10³ J = −5.48 kJ. The negative sign means work is done by the gas on the surroundings. Because the process is isothermal and the gas is ideal, ΔU = 0, so q = −w = +5.48 kJ of heat must be absorbed from the surroundings to keep the temperature constant.
Using the following data, calculate the standard enthalpy of formation of methane, CH₄(g): C(graphite) + O₂(g) → CO₂(g), ΔH° = −394 kJ/mol H₂(g) + ½O₂(g) → H₂O(l), ΔH° = −286 kJ/mol CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH° = −890 kJ/mol
Answer
The target equation is C(graphite) + 2H₂(g) → CH₄(g). By Hess's law, build the target from the given equations. Keep equation 1 as it is: C + O₂ → CO₂, ΔH = −394 kJ. Multiply equation 2 by 2: 2H₂ + O₂ → 2H₂O(l), ΔH = 2 × (−286) = −572 kJ. Reverse equation 3, which flips the sign: CO₂ + 2H₂O(l) → CH₄ + 2O₂, ΔH = +890 kJ. Adding the three: CO₂ and 2H₂O(l) cancel between the reactant and product sides, and 3O₂ on the left cancels 2O₂ on the right leaving O₂ balanced correctly, giving C(graphite) + 2H₂(g) → CH₄(g). ΔH°_f = −394 + (−572) + 890 = −76 kJ/mol. The negative value tells us methane is enthalpically more stable than its constituent elements taken separately.
For the hydrogenation reaction C₂H₄(g) + H₂(g) → C₂H₆(g), the average bond enthalpies (in kJ/mol) are C=C 615, C−C 348, C−H 413 and H−H 436. (i) Explain what a bond enthalpy is. (ii) List the bonds broken and the bonds formed. (iii) Calculate ΔH for the reaction. (iv) State whether the reaction is exothermic or endothermic and interpret the sign. (v) Explain why a value calculated this way may differ from the experimental enthalpy change.
Answer
(i) Bond enthalpy is the average enthalpy change required to break one mole of a particular type of bond in the gaseous state, separating the atoms in the gas phase. Breaking bonds always absorbs energy, so bond enthalpies are positive quantities, and forming bonds releases the same amount. (ii) In ethene the four C−H bonds are carried straight through into ethane unchanged, so only the reacting bonds need be counted. Bonds broken: one C=C and one H−H. Bonds formed: one C−C single bond and two new C−H bonds (the two hydrogen atoms attaching one to each carbon). (iii) Energy absorbed in breaking bonds = 615 + 436 = 1051 kJ. Energy released in forming bonds = 348 + (2 × 413) = 348 + 826 = 1174 kJ. ΔH = (bonds broken) − (bonds formed) = 1051 − 1174 = −123 kJ/mol. (iv) ΔH is negative, so the reaction is exothermic and releases 123 kJ for every mole of ethene hydrogenated. The bonds in the product are collectively stronger than those in the reactants, so the system falls to a lower enthalpy and the excess energy leaves as heat to the surroundings. This is why industrial hydrogenation vessels need cooling rather than heating once the catalyst starts working. (v) Bond enthalpies are averages taken over many different molecules containing that bond type. The actual strength of a C−H bond in ethane is not exactly the same as in methane or chloroform, since it depends on the neighbouring atoms. Bond enthalpy data also apply strictly to gaseous species, so any reaction involving liquids or solids brings in extra enthalpy of vaporisation or fusion terms. A calculation from average bond enthalpies is therefore a good estimate, typically within a few kJ, rather than an exact value; enthalpies of formation give more accurate answers.
A gas-phase reaction has ΔH° = +42.5 kJ/mol and ΔS° = +125 J K⁻¹ mol⁻¹, both assumed constant with temperature. (i) Calculate ΔG° at 298 K. (ii) State with reason whether the reaction is spontaneous at 298 K. (iii) Find the temperature at which the reaction is at equilibrium under standard conditions. (iv) Calculate the equilibrium constant K at 298 K. (v) Explain physically why raising the temperature helps this reaction. (R = 8.314 J K⁻¹ mol⁻¹)
Answer
(i) Convert ΔH° to joules: 42.5 kJ/mol = 42500 J/mol. ΔG° = ΔH° − TΔS° = 42500 − (298 × 125) = 42500 − 37250 = +5250 J/mol = +5.25 kJ/mol. (ii) ΔG° is positive, so at 298 K the forward reaction is non-spontaneous under standard conditions. The enthalpy cost of +42.5 kJ is not yet repaid by the entropy gain, because at this temperature the TΔS° term is only 37.25 kJ. (iii) At equilibrium ΔG° = 0, so ΔH° = TΔS°, giving T = ΔH° ÷ ΔS° = 42500 ÷ 125 = 340 K. Below 340 K the reaction is non-spontaneous and above 340 K it becomes spontaneous, so 340 K (about 67 °C) is the changeover temperature. (iv) Using ΔG° = −2.303 RT log K: log K = −ΔG° ÷ (2.303 RT) = −5250 ÷ (2.303 × 8.314 × 298) = −5250 ÷ 5705 = −0.920. K = antilog(−0.920) = 0.120. A value of K well below 1 is consistent with a positive ΔG°: at equilibrium the mixture at 298 K contains far more reactant than product. (v) The reaction absorbs heat but creates disorder, since ΔS° is positive — typically because more moles of gas are produced or a more mobile arrangement results. The entropy contribution to ΔG enters as −TΔS°, so its magnitude grows in direct proportion to temperature while ΔH° stays essentially fixed. Raising the temperature therefore increases the reward for creating disorder without increasing the enthalpy penalty, and past 340 K the entropy term overtakes it and ΔG turns negative. This is the thermodynamic reason endothermic decompositions such as limestone calcination are run in hot kilns.
A food technologist measures the energy content of a new snack bar. She burns 0.500 g of the dried sample (taken to have an average molar mass of 100 g/mol) in a bomb calorimeter of heat capacity 8.20 kJ/K, and records a temperature rise of 1.50 K. The combustion is represented as X(s) + 6O₂(g) → 5CO₂(g) + 5H₂O(l). Answer: (a) Calculate the heat released and hence ΔU per mole of X. (b) Explain why a bomb calorimeter gives ΔU and not ΔH. (c) Calculate ΔH per mole of X at 298 K. (d) Comment on why the difference between ΔU and ΔH is small here, and on one precaution needed for a reliable reading. (R = 8.314 J K⁻¹ mol⁻¹)
Answer
(a) Heat absorbed by the calorimeter q = C × ΔT = 8.20 kJ/K × 1.50 K = 12.3 kJ. This heat came out of the reacting sample, so the sample released 12.3 kJ. Moles of X burnt = 0.500 ÷ 100 = 5.00 × 10⁻³ mol. ΔU = −12.3 ÷ (5.00 × 10⁻³) = −2460 kJ/mol. The negative sign records that the system lost energy. (b) A bomb calorimeter is a rigid sealed steel vessel, so the volume of the reacting system cannot change. With ΔV = 0 the pressure–volume work w = −p_ext ΔV is zero, and the first law reduces to ΔU = q_v. The heat measured at constant volume is therefore the internal energy change directly. Enthalpy change equals the heat measured only at constant pressure, which would need an open or flexible container. (c) Only gases count in Δn_g. Products contain 5 mol of gaseous CO₂; the water is liquid and X is a solid. Reactants contain 6 mol of gaseous O₂. Δn_g = 5 − 6 = −1. ΔH = ΔU + Δn_g RT = −2460 + (−1)(8.314 × 10⁻³ kJ K⁻¹ mol⁻¹)(298 K) = −2460 − 2.48 = −2462.5 kJ/mol. (d) The correction is only about 2.5 kJ out of 2460 kJ, roughly 0.1%, because the RT term is small and Δn_g is just −1. Whenever a combustion produces and consumes nearly equal moles of gas, ΔU and ΔH are almost identical, which is why food energy values are quoted without fuss over which one was measured. For a reliable reading the calorimeter must first be calibrated with a substance of accurately known enthalpy of combustion so that its heat capacity is genuinely 8.20 kJ/K, the sample must be completely dried and fully burnt (any soot means incomplete combustion and a low result), and excess oxygen must be present at high pressure. The temperature rise should also be corrected for slow heat leakage to the surroundings.
RowQ generates fresh questions on Chemical Thermodynamics, marks your answers, and explains every step.
Start free