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Thermochemistry Questions

Thermochemistry is the study of heat and energy as they relate to physical transformations and chemical reactions. A reaction can absorb or release energy, and phase changes, such as boiling and melting, can occur. Thermochemistry is primarily concerned with energy changes, specifically the exchange of energy between a system and its surroundings.

Definition: Thermochemistry is the branch of thermodynamics that studies changes that occur during chemical reactions.

Thermochemistry Chemistry Questions with Solutions

Q-1: Given that standard molar enthalpies of formation of NO(g) and NO2(g) are respectively 90.3 kJ/mol and 33.2 kJ/mol. Calculate the enthalpy change for the reaction 2NO(g) + O2(g) →2NO2(g).

Answer: ΔH= ∑ΔHproducts– ∑ΔHreactants

ΔH= 2(33.2)-[2(90.3)+0]

ΔH= -114.2kJ

Note: Enthalpy of formation of free elements like O2 is always 0.

Q-2: The enthalpy of formation of AgCl is obtained from the enthalpy change from which one of the following process?

  1. Ag+(aq) +Cl(aq) →AgCl(s)
  2. Ag(s) + ½ Cl2(g) →AgCl(s)
  3. AgCl(s) →Ag(s) + ½ Cl2(g)
  4. Ag(s) +AuCl →Au(s) + AgCl(s)

Answer: b) Ag(s) + ½ Cl2(g) →AgCl(s)

Explanation: The enthalpy change accompanying the formation of one mole of a pure substance in its standard state from the constituents elements in their standard state is called enthalpy of formation. For Ag, the standard state is solid and for chlorine, it is gas. Thus, the process that gives the enthalpy of formation of AgCl is: Ag(s) + ½ Cl2(g) →AgCl(s)

Q-3: ΔH and ΔE for the reaction, Fe2O3 (s) +3H2(g) → 2Fe(s) + H2O(l) at constant temperature are related as

  1. ΔH = ΔE
  2. ΔH = ΔE +RT
  3. ΔH = ΔE +3RT
  4. ΔH = ΔE -3RT

Answer: d) ΔH = ΔE -3RT

Explanation: For any chemical reaction,

ΔH = ΔE + ΔngRT

Where, Δng = total number of moles of gaseous product- total number of moles of gaseous reactants.

For the given reaction,

Δng =0-3 = -3

ΔH = ΔE + (-3)RT

ΔH = ΔE -3RT

Q-4: For the reaction, N2(g) + 3H2(g) →2NH3 (g), compute the entropy change (in J/K/mol) for the process and comment on the sign of the property.

Data Species NH3(g) N2(g) H2(g)
So(J/K/mol) 192.3 191.5 130.6

Answer: ΔSo= ∑ΔSoproducts– ∑ΔSoreactants

ΔSo= 2×192.3-[191.5+3(130.6)] = -198.7J/K/mol

In the reaction, two gaseous species are there in the reactant side and the number of moles of gaseous species on the product side is 1. So there is a decrease in the number of gaseous species which is represented by the negative sign of entropy.

Q-5: The rate of evaporation of a liquid is always faster at a higher temperature because

  1. The enthalpy of vaporisation is always endothermic
  2. The enthalpy of vaporisation is always exothermic
  3. The enthalpy of vaporisation is zero
  4. The internal pressure of the liquid is less than that of the gas

Answer: a) The enthalpy of vaporisation is always endothermic

Explanation: At the time of evaporation, energy is required to overcome the intermolecular forces between the molecules of liquid, that is, the enthalpy of vaporisation is always endothermic. Since at high temperatures more energy is available, the rate of evaporation is faster at elevated temperatures.

Q-6: Given that

A(s) → A(l) ΔH= +x

A(l) → A(g) ΔH= +y

The heat of sublimation of A will be

a) x-y

b) x+y

c) x or y

d) -x+y

Answer: b) x+y

Explanation: It is a physical process by which a substance is directly converted from solid to gaseous.

The reaction for heat of sublimation is A(s) → A(g) which is obtained by adding A(s) → A(l) and A(l) → A(g). Thus, the heat of sublimation of A will be x+y.

Q-7: ΔH= -25kcal for the reaction CH4(g) +Cl2(g) → CH3Cl(g) + HCl(g).

Bond Bond Energy(kCal)
C-Cl 84
H-Cl 103
C-H x
Cl-Cl y
x:y 9:5

What is the bond enthalpy of Cl-Cl bond? Use the given data to calculate the answer.

Answer: ΔH= ∑ ΔB.Ereactants – ∑ ΔB.Eproducts

ΔH= 4(C-H)+Cl-Cl-[3(C-H)+1(C-Cl) +1(H-Cl)]

-25= 4x+y-[3x+84+103]

This implies, x+y-187= -25

Or x+y= 162

Given: x:y=9:5. Therefore, x= 9y/5

Substituting x= 9y/5 in x+y= 162 , we get, y= 57.85 kCal

Q-8: For the reaction,

N2(g) + 2O2(g) → 2NO2(g) ΔH = -66 kJ

Calculate the value of ΔfH of NO2.

Answer: Enthalpy of formation(ΔfH) is the enthalpy change accompanying the formation of one mole of a pure substance in its standard state from the constituent elements in their standard state.

The given ΔH of NO2 is for its two mole formation. Since we need to calculate the ΔfH of NO2 for one mole by definition, therefore it will be equal to

ΔfH = ΔH/2 = -66/2 = -33kJ

Q-9: Which of the following is not an endothermic reaction?

a) Decomposition of water

b) Conversion of graphite to diamond

c) Combustion of methane

d) Dehydrogenation of ethane to ethylene

Answer: c) Combustion of methane

Explanation: Heat is produced in any combustion reaction. The reactions that involve the release of heat are said to be exothermic.

Q-10: The enthalpy changes of the following reactions at 27oC are

1) Na(s) + ½ Cl2(g) → NaCl (s) ; ΔfH= -411kJ/mol

2) H2(g) + S(s) + 2O2(g) → H2SO4 (l) ; ΔfH= -811kJ/mol

3) 2Na(s) +S(s) + 2O2(g) → Na2SO4 (s) ; ΔfH= -1382kJ/mol

4) ½ H2(g) + ½ Cl2(g) → HCl (g) ; ΔfH= -92 kJ/mol

From the above data, the heat change of reaction at constant volume (in kJ/mol) at 27oC for the process

2NaCl(s) + H2SO4 (l) → Na2SO4 (s) + 2HCl (g) is:


Reverse equation 1 and multiply it by 2

a) 2NaCl (s)→ 2Na(s) + Cl2(g) ; ΔfH= +2×411kJ/mol = +822kJ/mol

Reverse equation 2

b) H2SO4 (l) →H2(g) + S(s) + 2O2(g) ; ΔfH= +811kJ/mol

c) 2Na(s) +S(s) + 2O2(g) → Na2SO4 (s) ; ΔfH= -1382kJ/mol

Multiply equation 4 by 2

d) H2(g) + Cl2(g) → 2HCl (g) ; ΔfH= -92×2= -184 kJ/mol

On adding equation, a, b,c,and d, we get

2NaCl(s) + H2SO4 (l) → Na2SO4 (s) + 2HCl (g); ΔfH

ΔfH= 822+811+(-1382-184) = 67kJ/mol

We know that, ΔfH = ΔU + ΔngRT

Where, Δng = total number of moles of gaseous product- total number of moles of gaseous reactants.

For the reaction given, Δng = 2

R= 8.3J/K mol = 0.0083kJ/mol

T= 27oC = 300K

ΔU is the heat change of reaction at constant volume

On substituting the values, we get

67= ΔU + 2×0.0083×300

ΔU= 62.02 kJ/mol

Q-11: Heat of combustion for benzene and acetylene are -3900 and -642 joule. Then calculate the heat of reaction(per mole) for the following reaction.

3C2H2→ C6H6


1) C6H6 + 15/2 O2 → 6CO2 + 3H2O

2) C2H2 + 5/2 O2 → 2CO2 + H2O

Multiply reaction 2 with 3 and reverse reaction 1 and add them, we get

3C2H2→ C6H6

ΔfH = 3(ΔHC)acetylene +(ΔHc)benzene = 3(-642)+ 3900 = 1974 J for 3 mole

Therefore, ΔfH for 1 mole= 1974/3 = 658J/mol

Q-12: For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, ΔCH= -601.70kJ/mol, the magnitude of change in internal energy for the reaction is _________ kJ. (Nearest integer)

Given: 8.3 J/Kmol

Answer: The combustion reaction for one mole of magnesium is

Mg(s) + ½ O2(g) → MgO(s)


Where, Δng = total number of moles of gaseous product- total number of moles of gaseous reactants.

ΔE is the internal energy

For the given reaction, Δng = -½

ΔE= -601.70 kJ/mol -½ (8.3×10-3 × 300) = 599.455 kJ = 600 kJ

Q-13: NaOH(s) has a heat of solution of -42.6kcal/mol NaOH. When NaOH is dissolved in water, the temperature of the solution

a) increases

b) decreases

c) remains constant

d) can’t be predicted

Answer: a) increases

Q-14: What is Hess’s Law?

Answer: Hess’s law is an important result of the first law of thermodynamics. It states that the enthalpy change in a chemical or physical process is the same whether it is performed in one step or several steps.

Q-15: If total enthalpies of reactants and products are HR and HP respectively, then for an endothermic reaction

a) HR<HP

b) HR>HP

c) HR=HP

d) Data insufficient

Answer: a) HR<HP

Explanation: For an endothermic reaction, enthalpy of reaction is positive. Enthalpy of reaction is calculated by using below equation:


For ΔH to be positive, HR must be less than HP.

Practise Questions on Thermochemistry

Q-1: Which of the following reactions does not represent ΔfH?

a) C(s) + O2 → CO2

b) Br2(l) + H2(g)→ 2HBr

c) CO + O2 → CO2

d) C(s)→ C(g)

Q-2: Which of the following pairs are correctly matched?

i) Arrhenius Equation Variation of enthalpy of a reaction with temperature
ii) Kirchhoff equation Variation of rate constant with temperature
iii) Second law of thermodynamics Entropy of an isolated system tends to increase and reach a maximum value
iv) Hess’s law of constant heat summation Enthalpy change in a reaction is always constant and independent of the manner in which the reaction occurs.

Q-3: From the following data at 25oC

Reaction ΔfHo (kJ/mol)
½ H2(g) + ½ O2(g)→ OH(g) 42
H2(g) + ½ O2(g)→ H2O(g) -242
H2(g)→2H(g) 436
O2(g)→2O(g) 495

Calculate the ΔfHo for the reaction H2O(g)→ 2H(g) + O(g)

Q-4: The heat of combustion of carbon is 394 kJ/mol. The heat evolved in combustion of 6.022×1022 atoms of carbon is:

Q-5: Calculate standard enthalpies of formation of CS2(l). Given the standard enthalpy of combustion of carbon (s), sulphur(s) and CS2(l) are: -393.3, -293.72 and -1108.76 kJ/mol respectively.

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