Thermochemistry & Structure — practice questions

Write the full electron configuration of a sulfur atom (Z = 16) and of a calcium atom (Z = 20).

Write the electron configurations of a nickel atom (Z = 28) and the Ni²⁺ ion, and explain which electrons are removed when the ion forms.

A student writes that 'the 4s electrons must be the hardest to remove from a titanium atom, because 4s was the last sublevel to be filled before 3d, and electrons are removed in the reverse order to the order in which they were added.' Evaluate this claim, and support your evaluation by writing the configurations of Ti (Z = 22), Ti²⁺ and Ti⁴⁺.

Define first ionisation energy, and state how it changes across a period from left to right.

Explain why the atomic radius of potassium is larger than the atomic radius of sodium, but the atomic radius of magnesium is smaller than that of sodium.

The ions O²⁻, F⁻, Na⁺ and Mg²⁺ all have the electron configuration 1s² 2s² 2p⁶. Their radii, in no particular order, are 0.065 nm, 0.140 nm, 0.095 nm and 0.133 nm. Assign each radius to its ion and fully justify your assignment. Compare and contrast your reasoning with the reasoning you would use to order the radii of the neutral atoms O, F, Na and Mg.

State the number of bonding pairs and lone pairs around the central atom in PF₅ and in NH₃, and name the shape of each.

The bond angle in CH₄ is 109.5°, in NH₃ it is 107°, and in H₂O it is 104.5°. Explain why the angle decreases across these three molecules.

Both XeF₄ and CF₄ contain four bonds to a central atom, yet XeF₄ is square planar and CF₄ is tetrahedral. Justify the difference in shape, and explain why a student who assumed 'four bonds always means tetrahedral' would also get the shape of ClF₃ wrong.

State whether each of CO₂, H₂O and CCl₄ is a polar or a non-polar molecule.

Both CO₂ and SO₂ contain two polar double bonds to oxygen, yet CO₂ is non-polar and SO₂ is polar. Explain this difference.

Consider XeF₂, XeF₄ and SF₄. Each has lone pairs on the central atom, yet only one of the three is polar. Identify the polar molecule and justify your answer fully, explaining why the presence of lone pairs does not by itself make a molecule polar.

Identify all the types of attractive force between the molecules in liquid methane, CH₄, and in liquid hydrogen fluoride, HF.

Explain why ethanol, CH₃CH₂OH, boils at 78 °C while ethanal, CH₃CHO, of very similar molar mass, boils at only 21 °C.

Propanone, CH₃COCH₃, boils at 56 °C and is completely miscible with water, yet propanone molecules cannot hydrogen bond to one another. Explain this apparent contradiction, and compare propanone with propan-1-ol (boiling point 97 °C), which has a similar molar mass.

Explain what must happen to the particles of a molecular substance when it boils, and state what is NOT broken.

The boiling points of the group 17 hydrides are HCl −85 °C, HBr −67 °C and HI −35 °C. Explain this trend.

Water (M = 18) boils at 100 °C while hydrogen sulfide (M = 34) boils at −60 °C. Ethanol (M = 46) boils at 78 °C while dimethyl ether, CH₃OCH₃ (M = 46), boils at −24 °C. Compare and contrast the reasons for these two large differences, and explain what the two comparisons together show about the relative importance of molecular size and hydrogen bonding.

State whether each of ammonia, iodine and methanol is likely to be soluble in water, and give the reason in each case.

Explain why methanol, CH₃OH, is completely miscible with water but octan-1-ol, CH₃(CH₂)₇OH, is essentially insoluble.

Propanone, CH₃COCH₃, is completely miscible with water. Butane, CH₃CH₂CH₂CH₃, of almost identical molar mass, is insoluble. Silver chloride, AgCl, is also insoluble in water despite being ionic. Justify all three observations, and explain why the reason for AgCl's insolubility is fundamentally different from the reason for butane's.

Define the standard enthalpy of formation, ΔfH°, and write the thermochemical equation that corresponds to ΔfH°(CO₂(g)) = −393.5 kJ mol⁻¹.

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), ΔrH° = −92.2 kJ mol⁻¹. Calculate the energy change when 5.60 g of nitrogen reacts completely with excess hydrogen, and explain why the answer is not −92.2 kJ. M(N₂) = 28.0 g mol⁻¹

A student states that 'ΔfH°(O₂(g)) must be negative, because forming the strong O=O bond releases energy.' Evaluate this statement. In your answer, explain what ΔfH° of an element in its standard state must be and why, and discuss why this convention is necessary for Hess's law calculations to work.

50.0 g of water is heated from 19.0 °C to 34.5 °C. Calculate the heat energy absorbed by the water, in kJ. c(water) = 4.18 J g⁻¹ °C⁻¹

Burning 0.780 g of ethanol, C₂H₅OH, raised the temperature of 200.0 g of water by 24.8 °C. Calculate the experimental enthalpy of combustion of ethanol. c(water) = 4.18 J g⁻¹ °C⁻¹, M(C₂H₅OH) = 46.0 g mol⁻¹

Two students measure the enthalpy of combustion of methanol with a spirit burner and a beaker of water. Student A obtains −520 kJ mol⁻¹ and Student B obtains −610 kJ mol⁻¹. The accepted value is −726 kJ mol⁻¹. Student A claims their result is closer to the truth because 'the smaller number means less heat was wasted'. Evaluate this claim, and discuss which experimental change would most improve both results.

ΔfusH°(I₂) = 15.5 kJ mol⁻¹ and ΔvapH°(I₂) = 41.6 kJ mol⁻¹. Calculate ΔsubH°(I₂) and explain why your answer is positive.

Calculate the energy required to melt 125 g of ice at 0 °C and then warm the resulting water to 35.0 °C. ΔfusH°(H₂O) = 6.01 kJ mol⁻¹, c(water) = 4.18 J g⁻¹ °C⁻¹, M(H₂O) = 18.0 g mol⁻¹

For water, ΔfusH° = 6.01 kJ mol⁻¹ and ΔvapH° = 40.7 kJ mol⁻¹. For methane, ΔfusH° = 0.94 kJ mol⁻¹ and ΔvapH° = 8.19 kJ mol⁻¹. Compare and contrast these two sets of values, explaining both why vaporisation costs so much more than fusion in each substance, and why both of water's values are so much larger than methane's.

State Hess's law, and explain why ΔfH° of O₂(g) is taken as zero in Hess's law calculations.

Calculate ΔrH° for the reaction 2NH₃(g) + 3N₂O(g) → 4N₂(g) + 3H₂O(l), given ΔfH°: NH₃(g) = −46.1, N₂O(g) = +82.1, H₂O(l) = −285.8 kJ mol⁻¹.

Use the data below to calculate ΔrH° for the reaction 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g). Explain why the ΔvapH° value is needed, and justify why Hess's law allows the answer to be obtained even though this reaction is difficult to study directly.

ΔfH°(NH₃(g)) = −46.1 kJ mol⁻¹; ΔfH°(NO(g)) = +90.3 kJ mol⁻¹; ΔfH°(H₂O(l)) = −285.8 kJ mol⁻¹; ΔvapH°(H₂O) = +44.0 kJ mol⁻¹

State whether ΔSsys is positive or negative for each of the following, and give a reason: (a) 2SO₂(g) + O₂(g) → 2SO₃(g); (b) NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq).

Explain why the reaction 2H₂(g) + O₂(g) → 2H₂O(l), ΔrH° = −572 kJ mol⁻¹, is spontaneous even though ΔSsys is negative.

A mixture of hydrogen and oxygen can be kept in a sealed container indefinitely at room temperature with no observable reaction, yet 2H₂(g) + O₂(g) → 2H₂O(l) has ΔrH° = −572 kJ mol⁻¹ and is spontaneous. Separately, the decomposition CaCO₃(s) → CaO(s) + CO₂(g) has ΔrH° = +178 kJ mol⁻¹ and does not occur at room temperature, but proceeds readily above about 900 °C. Fully explain both observations, and identify what is fundamentally different about the two cases.