Thermochemical Principles — practice questions

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.