CHEM*1050 Mid-Term Test #1 Winter 2026
Version C with Solutions
Following are the first midterm test questions for the Winter 2026 semester.
Included are the correct answers and the solutions. Different versions of the
exam had the questions and the accompanying multiple-choice options shuffled.
Be sure you are referring to the right exam version.
1. When ethylene, C2H4, evaporates at constant pressure, what are the signs
of the enthalpy change and the entropy change for the process?
A. ∆S > 0 and ∆H >0
B. ∆S > 0 and ∆H < 0
C. ∆S < 0 and ∆H > 0
D. ∆S < 0 and ∆H < 0
E. ∆S >0 and ∆H = 0
Evapouration always requires the input of energy so the liquid molecules can
break their bonds with each other and escape into the gas phase. So enthalpy
going into a system is defined as being positive. As for the entropy, there is a
liquid turning into a gas, so clearly the entropy must be increasing. Both ∆S
and ∆H are positive, and A is the answer.
2. An ideal gas is compressed isothermally (at constant temperature). From
the perspective of the system, the heat exchanged in the process is –120 J.
What is the value for the work?
A. +120 J
B. –120 J
C. 0 J
D. +240 J
E. –240 J
This is just a First Law of Thermodynamics problem. In an isothermal process,
the internal energy is unchanged. The question tells us that ∆U = q + w = 0.
So if the heat is –120 J, then the work must equal +120 J. The answer is A.
3. The standard enthalpy of formation of calcium chloride, CaCl2(s), is -795.4
kJ mol-1. All of the reactions below are found in the Born-Haber cycle for
CaCl2(s). Which reaction has an enthalpy change that is exothermic?
A. Ca(g) → Ca+(g)
B. Ca2+(g) + 2 Cl–(g) → CaCl2(s)
C. Ca(s) → Ca(g)
D. CaCl2(s) → Ca2+(g) + 2 Cl–(g)
E. CaCl2(s) → Ca(s) + Cl2(g)
Here you need to reason out which reaction releases enthalpy; the others
must absorb it to be able to occur. A requires the ejection of an electron.
Energy is required to kick that electron out of the atom. C is going from solid
to gas, the process of sublimation. This too requires energy to occur. D
involves the breaking apart of the solid into its constituent ions, the lattice
enthalpy. Separating the cations and anions clearly needs energy to overcome
the Coulomb attraction between them. So energy is again going in. By
contrast, B is the reverse of D, so it must be negative which makes the
process exothermic. It seems like B is the answer. Confirm that by checking
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, CHEM*1050 Mid-Term Test #1 Winter 2026
Version C with Solutions
on D which is the reverse of the formation reaction (with calcium and chlorine
in their elemental standard states). The question tells us that the formation
reaction is exothermic, so the reverse must be endothermic and positive.
Indeed the answer must be B.
4. Of the following reactions, which one will have a positive ∆S˚?
A. 2 CO(g) + O2(g) → 2 CO2(g)
B. 2 Na(s) + Cl2(g) → 2 NaCl(s)
C. H2(g) + Br2(𝓁) → 2 HBr(g)
D. 2 H2(g) + O2(g) → 2 H2O(g)
E. 3 O2(g) → 2 O3(g)
Since gases carry a lot of entropy compared to their liquid or solid
counterparts, we look for reactions that show an increase in the number of
gas phase species. Only answer C shows this (watch the phases closely – the
Br2 is in the liquid phase).
5. A model gas phase reaction, A(g) → B(g), has a Gibbs energy, ∆G, equal to 0 kJ
mol-1 under certain conditions. At the same time and under those same
conditions, ∆G˚ = –30 kJ mol-1. Which statement is true about this reaction at
these conditions?
A. The reaction will proceed in the forward direction to produce more B.
B. The reaction will proceed in the reverse direction to produce more A.
C. The reaction is at equilibrium and [A] > [B].
D. The reaction is at equilibrium and [A] = [B].
E. The reaction is at equilibrium and [A] < [B].
∆G˚ tells us about the location of equilibrium for a reaction. ∆G tells us in
which direction we need to move to reach equilibrium. In this case, ∆G = 0 so
that means we are at equilibrium. We have to choose between C, D, and E.
Recall that ∆G˚ = -RT lnK and K will have the form of [B]/[A]. For ∆G˚ to be
negative, lnK must be positive. That means that K >1. And that means that [A]
< [B]. The answer, therefore, is E.
6. Using the following bond enthalpy data (all in units of kJ mol-1: H-H 436;
C=C 611; C-H 414; C-C 347), determine an estimate of the molar enthalpy
of hydrogenation of ethylene
C H 2 =C H 2 \(g\) + H 2 \(g\) → C H 3 –C H 3 \(g\)
A. –128 kJ mol -1
B. –115 kJ mol-1
C. +128 kJ mol-1
D. +141 kJ mol-1
E. –141 kJ mol-1
The target reaction we are working toward is
C H 2 =C H 2 \(g\) + H 2 \(g\) → C H 3 -C H 3 \(g\)
Recall that when we use bond enthalpy data, it is the opposite of when we use
formation data. With BE, it is “reactants minus products”. Identify all of the
bonds involved. On the reactant side there is an H-H, a C=C bond, and 4 C-H
bonds. On the product side, we have one C-C bond and 6 C-H bonds. We
obtain
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Version C with Solutions
Following are the first midterm test questions for the Winter 2026 semester.
Included are the correct answers and the solutions. Different versions of the
exam had the questions and the accompanying multiple-choice options shuffled.
Be sure you are referring to the right exam version.
1. When ethylene, C2H4, evaporates at constant pressure, what are the signs
of the enthalpy change and the entropy change for the process?
A. ∆S > 0 and ∆H >0
B. ∆S > 0 and ∆H < 0
C. ∆S < 0 and ∆H > 0
D. ∆S < 0 and ∆H < 0
E. ∆S >0 and ∆H = 0
Evapouration always requires the input of energy so the liquid molecules can
break their bonds with each other and escape into the gas phase. So enthalpy
going into a system is defined as being positive. As for the entropy, there is a
liquid turning into a gas, so clearly the entropy must be increasing. Both ∆S
and ∆H are positive, and A is the answer.
2. An ideal gas is compressed isothermally (at constant temperature). From
the perspective of the system, the heat exchanged in the process is –120 J.
What is the value for the work?
A. +120 J
B. –120 J
C. 0 J
D. +240 J
E. –240 J
This is just a First Law of Thermodynamics problem. In an isothermal process,
the internal energy is unchanged. The question tells us that ∆U = q + w = 0.
So if the heat is –120 J, then the work must equal +120 J. The answer is A.
3. The standard enthalpy of formation of calcium chloride, CaCl2(s), is -795.4
kJ mol-1. All of the reactions below are found in the Born-Haber cycle for
CaCl2(s). Which reaction has an enthalpy change that is exothermic?
A. Ca(g) → Ca+(g)
B. Ca2+(g) + 2 Cl–(g) → CaCl2(s)
C. Ca(s) → Ca(g)
D. CaCl2(s) → Ca2+(g) + 2 Cl–(g)
E. CaCl2(s) → Ca(s) + Cl2(g)
Here you need to reason out which reaction releases enthalpy; the others
must absorb it to be able to occur. A requires the ejection of an electron.
Energy is required to kick that electron out of the atom. C is going from solid
to gas, the process of sublimation. This too requires energy to occur. D
involves the breaking apart of the solid into its constituent ions, the lattice
enthalpy. Separating the cations and anions clearly needs energy to overcome
the Coulomb attraction between them. So energy is again going in. By
contrast, B is the reverse of D, so it must be negative which makes the
process exothermic. It seems like B is the answer. Confirm that by checking
Page 1 of 10
, CHEM*1050 Mid-Term Test #1 Winter 2026
Version C with Solutions
on D which is the reverse of the formation reaction (with calcium and chlorine
in their elemental standard states). The question tells us that the formation
reaction is exothermic, so the reverse must be endothermic and positive.
Indeed the answer must be B.
4. Of the following reactions, which one will have a positive ∆S˚?
A. 2 CO(g) + O2(g) → 2 CO2(g)
B. 2 Na(s) + Cl2(g) → 2 NaCl(s)
C. H2(g) + Br2(𝓁) → 2 HBr(g)
D. 2 H2(g) + O2(g) → 2 H2O(g)
E. 3 O2(g) → 2 O3(g)
Since gases carry a lot of entropy compared to their liquid or solid
counterparts, we look for reactions that show an increase in the number of
gas phase species. Only answer C shows this (watch the phases closely – the
Br2 is in the liquid phase).
5. A model gas phase reaction, A(g) → B(g), has a Gibbs energy, ∆G, equal to 0 kJ
mol-1 under certain conditions. At the same time and under those same
conditions, ∆G˚ = –30 kJ mol-1. Which statement is true about this reaction at
these conditions?
A. The reaction will proceed in the forward direction to produce more B.
B. The reaction will proceed in the reverse direction to produce more A.
C. The reaction is at equilibrium and [A] > [B].
D. The reaction is at equilibrium and [A] = [B].
E. The reaction is at equilibrium and [A] < [B].
∆G˚ tells us about the location of equilibrium for a reaction. ∆G tells us in
which direction we need to move to reach equilibrium. In this case, ∆G = 0 so
that means we are at equilibrium. We have to choose between C, D, and E.
Recall that ∆G˚ = -RT lnK and K will have the form of [B]/[A]. For ∆G˚ to be
negative, lnK must be positive. That means that K >1. And that means that [A]
< [B]. The answer, therefore, is E.
6. Using the following bond enthalpy data (all in units of kJ mol-1: H-H 436;
C=C 611; C-H 414; C-C 347), determine an estimate of the molar enthalpy
of hydrogenation of ethylene
C H 2 =C H 2 \(g\) + H 2 \(g\) → C H 3 –C H 3 \(g\)
A. –128 kJ mol -1
B. –115 kJ mol-1
C. +128 kJ mol-1
D. +141 kJ mol-1
E. –141 kJ mol-1
The target reaction we are working toward is
C H 2 =C H 2 \(g\) + H 2 \(g\) → C H 3 -C H 3 \(g\)
Recall that when we use bond enthalpy data, it is the opposite of when we use
formation data. With BE, it is “reactants minus products”. Identify all of the
bonds involved. On the reactant side there is an H-H, a C=C bond, and 4 C-H
bonds. On the product side, we have one C-C bond and 6 C-H bonds. We
obtain
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