Thermochemistry MCQ Quiz - Objective Question with Answer for Thermochemistry - Download Free PDF

CONCEPT:

Enthalpy Change (ΔH) and Energy Diagram

• The enthalpy change (ΔH) of a reaction represents the difference in energy between the products and reactants.
• When ΔH is negative, the reaction is exothermic, meaning energy is released and the products have lower energy than the reactants.
• Energy diagrams for exothermic reactions show the reactants starting at a higher energy level and the products at a lower energy level, with an energy "drop" between them.

EXPLANATION:

• In the given reaction:

  C(s) + 2H₂(g) → CH₄(g); ΔH = -74.8 kJ mol^-1

• ΔH is negative, indicating that the reaction is exothermic.
• In an energy diagram for this reaction:
  • The reactants (C and H₂) start at a higher energy level.
  • The products (CH₄) are at a lower energy level, showing that energy is released during the reaction.
  • The difference in energy between the reactants and products corresponds to the magnitude of ΔH (74.8 kJ mol^-1).

Therefore, the correct energy diagram will show the reactants at a higher energy level than the products, with an energy drop of 74.8 kJ mol^-1.

CONCEPT:

Molar Heat Capacity

• Molar heat capacity is defined as the amount of heat required to raise the temperature of one mole of a substance by one Kelvin (or one degree Celsius).
• The molar heat capacity is expressed in units of Joules per mole per Kelvin (J/mol·K).
• For water (H₂O), the molar heat capacity is relatively high due to hydrogen bonding, which requires more energy to break and increase molecular motion.

EXPLANATION:

• The molar heat capacity of water is a well-known value: 75.3 J/mol·K.
• This value corresponds to Option 1.
• It is important to note that the specific heat capacity of water (per gram) is 4.184 J/g·K, which is different from the molar heat capacity.

Therefore, the molar heat capacity of water is 75.3 Joule per mol-Kelvin (Option 1).

CONCEPT:

Hess's Law of Heat Summation

• The heat of a reaction is the same whether it occurs in one step or a series of steps.
• Heat of formation (ΔHf) is the energy change when 1 mole of a substance in its standard state is formed from its pure elements in their standard states.
• Heat of combustion (ΔHc) is the energy change when 1 mole of a substance undergoes complete combustion in oxygen under standard conditions.

GIVEN DATA:

• Heat of formation of CO2(g), ΔHf[CO2(g)] = –393.5 kJ/mol
• Heat of formation of H2O(l), ΔHf[H2O(l)] = –286.0 kJ/mol
• Heat of combustion of benzene, ΔHc[C6H6] = –3267.0 kJ/mol
• Heat of formation of benzene, ΔHf[C6H6], is to be calculated

Explanation:-

• The combustion reaction for benzene is:
  • C6H6(l) + 15/2 O2(g) ⟶ 6 CO2(g) + 3 H2O(l)

ΔH_f[CO₂(g)] = –393.5 kJ / mole

ΔH_f[H₂O(ℓ)] = –286.0 kJ / mole

ΔH_c[C₆H₆] = –3267.0 kJ / mole

ΔH_f C₆H₆ = (?)

\(\mathrm{C}_{6} \mathrm{H}_{6}+\frac{15}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 6 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\ell)\)

ΔH_R = ΔH_C = ∑ΔH_f(P) – ∑ΔH_f(R)

–3267 = 6 × (–393.5) + 3(–286) – ΔH_f(C₆H₆)

ΔH_f (C₆H₆) = 48 kJ/mole 

Conclusion:-

So, The heat of formation of benzene is 48 kJ mol–1 

CONCEPT:

Heat of Formation and Reaction Energies

• The heat of formation (ΔHf) of a compound is the enthalpy change when one mole of a compound is formed from its elements in their standard states.
• The heat of a reaction (ΔHr) can be related to the heats of formation of reactants and products.

Explantion:-

\(\underset{(\mathrm{g})}{\mathrm{SO}_{2}}+\frac{1}{2} \mathrm{O}_{(\mathrm{g})} \longrightarrow \underset{(\mathrm{g})}{\mathrm{SO}_{3}} \quad \Delta \mathrm{H}=-\mathrm{y}\)

\(\Delta \mathrm{H}_{\mathrm{r}}=\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_{3}}-\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_{2}}\)

\(-\mathrm{y}=-2 \mathrm{x}-\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_{2}}\)

\(\left(\Delta \mathrm{H}_{\mathrm{f}}\right)_{\mathrm{SO}_{2}}=\mathrm{y}-2 \mathrm{x}\)

CONCLUSION:

The correct answer is: y - 2x kcal.

CONCEPT:

Enthalpy of Neutralization

• The enthalpy of neutralization is the heat change when one mole of water is formed from the neutralization of an acid with an alkali.
• This heat change is constant for strong acids and strong bases and is typically -57.3 kJ/mol.
• For weak acids, less energy is released due to the incomplete ionization of the acid in solution.

EXPLANATION:

• Comparing the given options:
  • (1) 30 mL HCl and 30 mL NaOH
    • HCl is a strong acid, and NaOH is a strong base.
    • moles of acid or base = 30 mM
    • Neutralization is complete, leading to the highest temperature increase.
  • (2) 30 mL CH₃COOH and 30 mL NaOH
    • CH₃COOH is a weak acid, and NaOH is a strong base.
    • moles of acid or base = 30 mM
    • Due to the weak acid, less energy is released compared to strong acid-base neutralization.
  • (3) 50 mL HCl and 20 mL NaOH
    • HCl is a strong acid, and NaOH is a strong base.
    • moles of acid or base = 20 mM (limiting reagent)
    • Neutralization is incomplete, leading to a lower temperature increase.
  • (4) 45 mL CH₃COOH and 25 mL NaOH
    • CH₃COOH is a weak acid, and NaOH is a strong base.
    • moles of acid or base = 25 mM (limiting reagent)
    • Due to the weak acid and incomplete neutralization, less energy is released.

Therefore, the correct answer is 30 mL HCl and 30 mL NaOH (Option 1).

Explanation and Calculation:

• Bond Dissociation Energies (approximate values):
  • C-H bond inC2H6 : 413 kcal/mol
  • C-C bond in C2H6 : 348 kcal/mol
  • H-H bond: 436 kcal/mol
  • C-H bond in CH₄ : 413 kcal/mol
• Total Bond Dissociation Energy:
  • For C2H6 :
    • 1 x 348 + 6 x 413 = 2826 kcal
  • For H₂ :
    • 1 x 436 = 436 kcal
  • Total energy to break bonds:
    • 3262 kcal
• Total Bond Formation Energy:
  • For 2 CH₄ :
    • 2 x (4 x 413) = 3304 kcal
  • Total energy to form bonds:
    • 3304 kcal
• Heat Change ( \Delta H ) for the reaction:
  • \(\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed} \\ \Delta H = 3262 \text{ kcal} - 3304 \text{ kcal} = -42 \text{ kcal}\)
  • Convert to calories:
    • \(\Delta H = -42000 \text{ cal}\)

Conclusion:

The heat change for this reaction is closest to 41745 cal

Concept :

• For bond breaking heat is required.
• For bond-forming energy generated.
• Net energy change is the enthalpy of formation.
• Enthalpy of energy is the energy change due to the formation of one mole of the substance.
• Exothermic reactions are those reactions for which energy is generated.
• For exothermic reaction enthalpy of formation is negative.
• Endothermic reactions are those for which energy is absorbed.
• For endothermic reaction enthalpy of formation is positive.

Explanation:

• A reaction may be exothermic or endothermic. 
• If the energy released due to bond formation is greater than the energy required for bond breaking then the reaction is exothermic.
• For exothermic reaction enthalpy of formation is positive.
• If the energy released due to bond formation is less than the energy required for bond breaking then the reaction is endothermic.
• For endothermic reaction enthalpy of formation is positive.

Conclusion :

As the reaction may be endothermic or exothermic 

Then the enthalpy of formation of the reaction may be positive and negative 

So, the correct option is 3.

Concept:

• Entropy refers to the randomness present in the system.
• The value of randomness can be positive or negative, depending on the resultant of the system's randomness.
• The order of randomness is (maximum) Gas > liquid > solid (minimum).

• As, when solid converts to its liquid or gaseous state, the entropy increases, as molecules of the solid-state are packed tightly because of more force of attraction between the molecules, but as it is converted to liquid or the gaseous state, the force of attraction reduces and therefore, the entropy increases.

• If the state on both sides of the reaction, i.e., reactant and product, are the same, the entropy value can be determined by calculating the value of Δng, which refers to the change in the moles of gas molecules.

Δng = moles of gas in the product – no. of moles of gas in the reactant.

Explanation:

From the question,

• H₂O(l) \(\rightleftharpoons\) H₂O (v), ΔS > 0
• Expansion of gas at a constant temperature, ΔS > 0
• Sublimation of solid to gas, ΔS > 0
• 2H(g) → H₂(g), ΔS < 0 (∵ Δng < 0)

Hence, the correct option for the given question is, 4) 2H(g) → H2(g).

CONCEPT:

Enthalpy Change (ΔH) and Energy Diagram

• The enthalpy change (ΔH) of a reaction represents the difference in energy between the products and reactants.
• When ΔH is negative, the reaction is exothermic, meaning energy is released and the products have lower energy than the reactants.
• Energy diagrams for exothermic reactions show the reactants starting at a higher energy level and the products at a lower energy level, with an energy "drop" between them.

EXPLANATION:

• In the given reaction:

  C(s) + 2H₂(g) → CH₄(g); ΔH = -74.8 kJ mol^-1

• ΔH is negative, indicating that the reaction is exothermic.
• In an energy diagram for this reaction:
  • The reactants (C and H₂) start at a higher energy level.
  • The products (CH₄) are at a lower energy level, showing that energy is released during the reaction.
  • The difference in energy between the reactants and products corresponds to the magnitude of ΔH (74.8 kJ mol^-1).

Therefore, the correct energy diagram will show the reactants at a higher energy level than the products, with an energy drop of 74.8 kJ mol^-1.

CONCEPT:

Heat of Neutralisation

• The heat of neutralization is the energy change when an acid and a base react to form one mole of water in a neutralization reaction.
• For strong acid and strong base reactions, the heat of neutralization is typically high and nearly constant, around \(-57.1 \, kJ/mol\) because the reaction involves the direct combination of \(H^+\) and \(OH^−\) ions.
• For reactions involving weak acids or bases, the heat of neutralization is lower because part of the energy is used to ionize the weak acid or base.

EXPLANATION:

• Option 1: NaOH and CH₃COOH - Involves a strong base (NaOH) and a weak acid (CH₃COOH), so the heat of neutralization is lower.
• Option 2: HCl and NH₄OH - Involves a strong acid (HCl) and a weak base (NH₄OH), so the heat of neutralization is lower.
• Option 3: NH₄OH and CH₃COOH - Involves both a weak acid and a weak base, resulting in even lower heat of neutralization.
• Option 4: NaOH and HCl - Involves a strong acid (HCl) and a strong base (NaOH), leading to the maximum heat of neutralization, which is around \(-57.1 \, kJ/mol\).

The correct answer is (4) NaOH and HCl.

Concept :

• Sublimation: Sublimation is the process where the gaseous phase is formed directly from the solid phase of a substance without going through the liquid state. 
  • Enthalpy of sublimation is the heat required to sublimate one mole of substances at a temperature and pressure.
• Fusion: Fusion is the process where a liquid state is formed directly from the solid phase.
  • Enthalpy of fusion is the heat required to make solid to liquid 
• Vaporization: Vaporization is the process to make a liquid state into a gaseous state of a substance 
  • Enthalpy of vaporization is the heat required to make liquid vapor 
• Hess's law: according to Hess's law a chemical reaction net heat change of a chemical reaction will be the same if the reaction is carried out in a single step or by multiple steps.

Explanation : 

solid  liquid  H1,

Where H1 is the enthalpy of fusion t

liquid  gas  H2

Where H2 is the enthalpy of vaporization

Then by adding the two equation 

solid  gas  H3

Where H3 is the enthalpy of sublimation 

Then according to Hess's law 

H3 = H1 + H2

Enthalpy of sublimation = enthalpy of fusion + enthalpy of vaporization

Conclusion : 

So, we get 

Enthalpy of sublimation = enthalpy of fusion + enthalpy of vaporization

Thus, the correct option is (1).

Correct answer: 4)

Concept:

• Chemical reactions that absorb (or use) energy overall are called endothermic. In endothermic reactions, more energy is absorbed when the bonds in the reactants are broken than is released when new bonds are formed in the products.
• Endothermic reactions are accompanied by a decrease in the temperature of the reaction mixture.
• Chemical reactions that release energy are called exothermic. In exothermic reactions, more energy is released when the bonds are formed in the products that are used to break the bonds in the reactants.
• Exothermic reactions are accompanied by an increase in the temperature of the reaction mixture.

Explanation:

• Carbon reacts with Sulphur to give carbon disulphide, which is an endothermic reaction, where 92kJ/mol of heat is absorbed.
• Since the sum of F-F and O=O bond energy is much greater than the F-O-F bond.
• ΔH  is positive for the reaction.
• So, this is an endothermic reaction.
• In the case of \(\rm N_2+\frac{1}{2}O_2\rightarrow N_2O\), when N2 reacts with O2, heat is absorbed as the sum of N-N and O=O bond energy is much greater than the N-O-N bond.
• In case of \(\rm H_2+\frac{1}{2}O_2\rightarrow H_2O\)
• H₂(g) + 1/2O₂(g) → H₂O(g), ΔH = -57.82 kcal
• heat is released so this is an exothermic process. 

Conclusion:

 Thus, \(\rm H_2+\frac{1}{2}O_2\rightarrow H_2O\) is not an endothermic process.

Correct answer: 4)

Concept:

• The amount of heat energy produced on the complete combustion of 1 kg of a fuel is called its calorific value.
• The calorific value of a fuel is expressed in a unit called kilojoule per kg (kJ/kg).
• It is the variable of the heat or energy released, which is either measured in Gross Calorific Value (GCV) or Net Calorific Value (NCV).
• The calorific value of a fuel can be determined using a bomb calorimeter.

Explanation:

Given, m = 1.0 kg = 1000 g, specific heat of H2O (c) = 1 cal/g

\(\Delta T\) = 50-10 = 40 °C

Heat is supplied to raise the temperature of the water, 

Q= m. c. \(\Delta T\)

Q = 1000 x 1 x 40

Q= 40 x 10³

Calorific value = \(\frac{\Delta H}{Weight\: of \: fuel}\)

Calorific value = \(\frac{40\times 10^{3}}{10}\)

Calorific value = 4.0 x 10³ cal

Calorific value = 4000 cal

Conclusion:

Thus, the fuel value of coke is 4000 cal.

CONCEPT:

Enthalpy Change of Combustion (ΔH_comb)

• The enthalpy change of combustion is the heat energy released when one mole of a substance is burned completely in oxygen under standard conditions.
• The enthalpy change for a reaction can be calculated using the enthalpies of formation of the reactants and products:

  ΔH_reaction = Σ ΔH_f,products - Σ ΔH_f,reactants

EXPLANATION:

• Given the combustion reaction of ethylene (C₂H₄):

  C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l)

• Using the given standard enthalpies of formation:
  • ΔH_f[C₂H₄(g)] = 52 kJ/mol
  • ΔH_f[CO₂(g)] = -394 kJ/mol
  • ΔH_f[H₂O(l)] = -286 kJ/mol
• Calculate the enthalpy change for the combustion reaction:
  • ΔH_reaction = [2(ΔH_f[CO₂(g)]) + 2(ΔH_f[H₂O(l)])] - [ΔH_f[C₂H₄(g)] + 3(ΔH_f[O₂(g)])]
  • Since ΔH_f[O₂(g)] = 0 (standard state), the equation simplifies to:
  • ΔH_reaction = [2(-394) + 2(-286)] - [52 + 0]
  • = [-788 - 572] - 52
  • = -1360 - 52
  • = -1412 kJ/mol

Therefore, the change in enthalpy for the combustion of C₂H₄ is -1412 kJ/mol.

Concept:

Entropy of vapourisation - It is the entropy change when 1 mole of liquid changes into vapour at its boiling temperature.

If ΔS is change in entropy, then 

ΔS = S_vap - S_liq =   \(\frac{Δ H_{vap}}{T}\)

where,

• \(Δ H_{vap}\) = Enthalpy pof vapourisation per mole
• Sliq = molar entropy of the liquid
• Svap = molar entropy of the vapour
• T = boiling temperature in Kelvin

Gibbs Free Energy equation :- ΔG = ΔH - TΔS 

Calculation:

At 1 atmospheric pressure, the system is at equilibrium. Free energy associated with vapours is equal to zero.

∴  ΔG = 0

And the Gibbs free energy equation becomes,

ΔH - TΔS = 0

or

ΔH = TΔS

or T = \(\frac{\Delta H}{\Delta S}\)  , T is the temperature

Given, 

ΔHvap = 30 kJ/mol  = 30 × 10³ J\mol

ΔSvap = 75 mol-1 k-1

∴ T =  \(\frac{30\times 10^3}{75}\) = 400 K

Conclusion:

Therefore, the temperature of vapour at one atmosphere is 400K.

Hence, the correct answer is option 1.