MDCAT Chemistry chapter 8: Chemistry · Thermochemistry and energetics MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
32 exam-style questions on Chemistry · Thermochemistry and energetics, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Ten Chemistry · Thermochemistry and energetics questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. A gas in a cylinder is compressed: 40 kJ of work is done on it while it gives out 15 kJ of heat to the surroundings. Using ΔE = q + w, the change in its internal energy is
A
−25 kJ
B
+55 kJ
C
+25 kJ
D
−55 kJ
Show answer
Answer: C. Heat leaves the system, so q = −15 kJ; work is done on the system, so w = +40 kJ. ΔE = −15 + 40 = +25 kJ. Adding the two magnitudes gives +55 kJ, and reversing both signs gives −25 kJ.
2. The standard enthalpies of combustion of graphite and diamond are −393.5 and −395.4 kJ mol−1; both burn to CO2(g). For C(graphite) → C(diamond), ΔH° and the correct conclusion are
A
+1.9 kJ mol−1; diamond is higher in enthalpy than graphite
B
−1.9 kJ mol−1; diamond is lower in enthalpy than graphite
C
+1.9 kJ mol−1; graphite is higher in enthalpy than diamond
D
−788.9 kJ mol−1; diamond is lower in enthalpy than graphite
Show answer
Answer: A. By Hess's law, ΔH° = ΔH°c(graphite) − ΔH°c(diamond) = −393.5 + 395.4 = +1.9 kJ mol−1, so diamond lies higher and gives out more heat when it burns. This is why graphite, not diamond, is the standard state of carbon with ΔH°f = 0; adding the two combustion values gives −788.9.
3. The standard enthalpy of formation of carbon monoxide is found indirectly with Hess's law, not by burning carbon in a limited supply of oxygen. Why?
A
the product is always a mixture of CO and CO2
B
the formation of CO from carbon is endothermic
C
carbon monoxide does not burn in oxygen
D
enthalpy changes of formation cannot be measured
Show answer
Answer: A. Some carbon always burns on to CO2, so the heat measured does not belong to C + ½O2 → CO alone. Instead ΔH°f(CO) = ΔH°c(graphite) − ΔH°c(CO) = −393.5 − (−283.0) = −110.5 kJ mol−1.
4. Which statement about the signs of the enthalpy changes involved in dissolving an ionic solid in water is correct?
A
enthalpy of hydration of an ion is always positive; enthalpy of solution is always negative
B
enthalpy of hydration of an ion is always negative; enthalpy of solution may have either sign
C
enthalpy of hydration of an ion and enthalpy of solution are both always negative
D
enthalpy of hydration of an ion may have either sign; enthalpy of solution is always positive
Show answer
Answer: B. Water molecules are attracted to a gaseous ion and cluster round it, so hydration always releases energy. Dissolving a solid also needs energy to separate its ions, so the enthalpy of solution may be negative (NaOH warms the water) or positive (NH4NO3 cools it).
5. Using the bond energies H–H 436, Cl–Cl 242 and H–Cl 431 kJ mol−1, ΔH for H2(g) + Cl2(g) → 2HCl(g) is
A
+184 kJ
B
−184 kJ
C
+247 kJ
D
−862 kJ
Show answer
Answer: B. ΔH = bonds broken − bonds formed = (436 + 242) − 2 × 431 = 678 − 862 = −184 kJ. Counting only one H–Cl bond gives +247 kJ, and subtracting the other way round gives +184 kJ.
6. A calorimeter has a heat capacity of 800 J K−1. A reaction in it raises the temperature by 5.0 K. How much heat did the calorimeter absorb?
A
4.0 kJ
B
0.16 kJ
C
4000 kJ
D
800 kJ
Show answer
Answer: A. Heat absorbed = heat capacity × temperature rise = 800 J K−1 × 5.0 K = 4000 J = 4.0 kJ. Dividing 800 by 5.0 gives the wrong quantity (0.16 kJ), and 4000 kJ forgets that the answer is in joules.
7. At constant pressure ΔH = ΔE + PΔV. For which reaction, carried out at constant pressure, is ΔH greater than ΔE?
A
CaCO3(s) → CaO(s) + CO2(g)
B
N2(g) + 3H2(g) → 2NH3(g)
C
H2(g) + I2(g) → 2HI(g)
D
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Show answer
Answer: A. PΔV is positive only when the system expands, that is, when the moles of gas increase. Decomposing CaCO3 forms 1 mol of gas from none, so ΔH > ΔE. H2 + I2 has no change in gas moles (ΔH = ΔE), and the other two reactions reduce the gas moles (ΔH < ΔE).
8. A reaction takes place in a sealed, rigid steel vessel, and 12 kJ of heat flows out of the reaction mixture to the surroundings. For the reaction mixture,
A
w = 0 and ΔE = +12 kJ
B
w = −12 kJ and ΔE = 0
C
w = +12 kJ and ΔE = 0
D
w = 0 and ΔE = −12 kJ
Show answer
Answer: D. The volume cannot change, so no pressure–volume work is done (w = 0) and ΔE = qv. Heat leaves the system, so q = −12 kJ and the internal energy falls by exactly that amount: ΔE = −12 kJ.
9. Which statement correctly defines enthalpy, H, and relates its change to heat?
A
H = E − PV; ΔH equals the heat absorbed at constant pressure
B
H = E + PV; ΔH equals the heat absorbed at constant pressure
C
H = E + PV; ΔH equals the heat absorbed at constant volume
D
H = q + w; ΔH equals the heat absorbed at constant volume
Show answer
Answer: B. Enthalpy is defined as H = E + PV. At constant pressure ΔH = ΔE + PΔV = qp, so ΔH is the heat exchanged at constant pressure; the heat exchanged at constant volume equals ΔE. The expression q + w gives ΔE, not H.
10. Which statement is the first law of thermodynamics?
A
the internal energy of a system stays constant in every process at all times
B
heat flows by itself only from a hotter body to a colder body
C
energy can be neither created nor destroyed, only changed in form
D
all spontaneous processes give out heat to their surroundings
Show answer
Answer: C. The first law is the law of conservation of energy: the total energy of a system and its surroundings is constant, so ΔE = q + w. The internal energy of the system itself can change, and many spontaneous changes are endothermic.
Chemistry · Thermochemistry and energetics: questions students ask
How many Chemistry · Thermochemistry and energetics MCQs are there?
32, each with the answer and a short explanation, across 2 topics: Enthalpy changes, First law and Hess's law. Every chapter test draws ten of them, so you can take it many times.
Is the MDCAT Chemistry chapter 8 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
