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MCAT · Chemical and Physical Foundations · General Chemistry
Thermochemistry and kinetics is part of General Chemistry in the MCAT Chemical and Physical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.
Question types you will see here include calorimetry with q = mc delta T, hess's law and bond enthalpies, gibbs free energy and spontaneity and rate law from initial rates.
Common traps: using the wrong sign for the reaction compared with the surroundings; using reagent mass rather than solution mass; forgetting to scale to per mole.
Question 1
How much heat is needed to raise the temperature of 50 g of water by 20 °C? (Specific heat of water ≈ 4.2 J/(g·°C).)
4.2 J
238.095 J
1,000 J
4,200 J
Answer: D. 4,200 J
q = mcΔT = (50)(4.2)(20) = 4,200 J. Leaving out the specific heat gives 1000 J.
Question 2
How much heat is required to warm 2.0 kg of intravenous fluid (specific heat capacity 4.2 kJ/(kg·K)) from 20 °C to 37 °C?
14 kJ
71 kJ
140 kJ
310 kJ
Answer: C. 140 kJ
q = mcΔT = 2.0 kg × 4.2 kJ/(kg·K) × 17 K ≈ 143 kJ; a kelvin and a Celsius degree are the same size, so ΔT = 17. The 71 kJ choice forgets the mass, 310 kJ uses the final temperature of 37 instead of the 17 K change, and 14 kJ misplaces a decimal.
Question 3
A reaction has ΔH > 0 and ΔS > 0. Which statement describes its spontaneity under standard conditions?
It is spontaneous at all temperatures.
It is nonspontaneous at all temperatures.
It is spontaneous only at sufficiently high temperatures.
It is spontaneous only at sufficiently low temperatures.
Answer: C. It is spontaneous only at sufficiently high temperatures.
With ΔG = ΔH − TΔS, a positive ΔH is overcome only when TΔS is large enough, meaning T > ΔH/ΔS. Spontaneity only at low temperature applies to the opposite case, where ΔH < 0 and ΔS < 0.
Question 4
A first-order drug degradation has a half-life of 10 min. What fraction of the original drug remains after 50 min?
1/32
1/16
1/10
1/5
Answer: A. 1/32
50 min is five half-lives, so (1/2)⁵ = 1/32 remains (A). B counts four half-lives, C treats the decay as linear in time, and D divides 1 by the number of half-lives.
Question 5
For an exothermic, reversible reaction at equilibrium, raising the temperature will:
increase both rate constants and decrease the equilibrium constant.
decrease both rate constants and increase the equilibrium constant.
increase only the forward rate constant and increase the equilibrium constant.
increase both rate constants and leave the equilibrium constant unchanged.
Answer: A. increase both rate constants and decrease the equilibrium constant.
By the Arrhenius equation every rate constant increases with temperature, but the reverse (endothermic) direction has the larger activation energy and speeds up more, so K decreases, consistent with Le Châtelier's principle treating heat as a product. Saying K is unchanged confuses a temperature change with adding a catalyst.
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