Thermochemistry MCQ Quiz in मल्याळम - Objective Question with Answer for Thermochemistry - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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:

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:

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).

Concept:

The bond enthalpy, also known as bond dissociation energy, is the energy required to break a bond in a molecule and separate the atoms in their gaseous states. Diborane (B₂H₆) features a unique type of bonding called the "three-center two-electron bond," which involves three atoms sharing a pair of electrons. This bond is characterized differently compared to traditional two-center bonds.

Explanation: 

Given the following data:

• ΔfH° BH₃(g) = 100 kJ/mol (Enthalpy of formation of BH₃ gas)
• ΔfH° B₂H₆(g) = 36 kJ/mol (Enthalpy of formation of B₂H₆ gas)
• ΔH_atomization B(s) = 565 kJ/mol (Enthalpy of atomization of solid boron)
• ΔH_atomization H₂(g) = 436 kJ/mol (Enthalpy of atomization of hydrogen gas)

We derive the bond enthalpy of the three-center two-electron bond in B2H6.

• Starting with the dissociation process for BH₃(g):

  \( \text{B(s)} + \frac{3}{2}\text{H₂(g)} \rightarrow \text{BH₃(g)}, Δ_fH° [BH₃(g)] = 100 \text{ kJ/mol}\)

  Using the enthalpy values given:

  • \(BE(\text{B-H}) = 100 - 565 + \frac{3}{2} \times 436 \\BE(\text{B-H}) = 100 - 565 + 654\\BE(\text{B-H}) = 373 \text{ kJ/mol}\)

• Next, considering the formation of B2H6(g):

  \(2\text{B(s)} + 3\text{H₂(g)} \rightarrow \text{B₂H₆(g)}, \quad {Δ_fH° [B₂H₆(g)]} = 36 \text{ kJ/mol}\)

  Using the given data and bond enthalpies:

• \(ΔH = 2 \times 565 + 3 \times 436 - \{4 \times 373 + 2 \times BE(\text{3c-2e bond})\} = 36\)

• \(1130 + 1308 - \{1492 + 2 \times BE(\text{3c-2e bond})\} = 36\)

• \(2438 - 1492 - 2 \times BE(\text{3c-2e bond}) = 36\)

• \(946 = 36 + 2 \times BE(\text{3c-2e bond})\)

• \(2 \times BE(\text{3c-2e bond}) = 910 \)

• \(BE(\text{3c-2e bond}) = 455 \text{ kJ/mol}\)

Conclusion:

The bond enthalpy of the three-center two-electron bond in B₂H₆ is approximately 455 kJ/mol.

Concept:

To determine the enthalpy of dehydration of blue vitriol (Copper (II) sulphate pentahydrate), we use the enthalpy changes of related processes:

• Enthalpy of Solution (ΔH_sol): The change in enthalpy when one mole of a substance dissolves in a solvent.
  • For anhydrous CuSO₄: ΔH_sol = -66.11 kJ/mol
  • For CuSO₄•5H₂O: ΔH_sol = +11.5 kJ/mol (since heat is absorbed)
• Enthalpy of Dehydration: The enthalpy change when water is removed from a hydrated compound.
• Hess's Law: The total enthalpy change of a reaction is the sum of the enthalpy changes of the steps into which the reaction can be divided.

Explanation:

1. Step 1 - Dissolution of CuSO₄•5H₂O:
   • CuSO₄•5H₂O (s) → Cu²⁺ (aq) + SO₄^2- (aq)
   • ΔH_{sol, CuSO4•5H2O} = +11.5 kJ/mol
2. Step 2 - Dissolution of anhydrous CuSO₄:
   • CuSO₄ (s) → Cu²⁺ (aq) + SO₄^2- (aq)
   • ΔH_{sol, CuSO4} = -66.11 kJ/mol
3. Step 3 - Subtracting the two processes to find the enthalpy of dehydration:
   • ΔH_dehydration = ΔH_{sol, CuSO4•5H2O} - ΔH_{sol, CuSO4}
   • ΔH_dehydration = 11.5 kJ/mol - (-66.11 kJ/mol)
   • ΔH_dehydration = 11.5 kJ/mol + 66.11 kJ/mol
   • ΔH_dehydration = 77.61 kJ/mol

Conclusion:

Therefore, the enthalpy of dehydration of blue vitriol (Copper (II) sulphate pentahydrate) is +77.61 kJ/mol. Hence, the correct answer is: +77.61 kJ

Concept:

1. Heat of Reaction

• Also known as enthalpy change (ΔH) of the reaction.

• It is the amount of heat absorbed or released during a chemical reaction.

• Depends on the nature of reactants and products and the reaction conditions (temperature, pressure).

• Can be exothermic (heat released) or endothermic (heat absorbed).

• Expressed in kJ/mol for the stoichiometric amounts of reactants.

2. Heat of Combustion

• Amount of heat released when one mole of a substance burns completely in oxygen.

• Involves the formation of common combustion products, typically CO₂ and H₂O for organics.

• Represents the total energy released as chemical bonds are broken and formed during combustion.

• Standard heat of combustion is measured under standard conditions (25°C, 1 atm).

• Used to determine the energy content of fuels.

3. Heat of Formation

• Amount of heat change when one mole of a compound is formed from its elements in their standard states.

• Referred to as standard enthalpy of formation (ΔH_f°).

• Standard states are the most stable forms of the elements at 25°C and 1 atm.

• Used in Hess's Law to calculate the overall enthalpy change of a reaction.

• Values are tabulated for many compounds and used in thermodynamic calculations.

Concept:

The standard enthalpy change of a reaction (ΔH_reaction) can be calculated using the standard enthalpies of formation (ΔH_f⁰) of the reactants and products. The equation for the standard enthalpy change is:

ΔH_reaction = Σ ΔH_f⁰ (products) - Σ ΔH_f⁰ (reactants)

Explanation:

Given Data:

• ΔH_f⁰ of CO₂(g): -393 kJ/mol

• ΔH_f⁰ of CO(g): -110 kJ/mol

• ΔH_f⁰ of N₂O(g): 819 kJ/mol

• ΔH_f⁰ of NO₂(g): 34 kJ/mol

Reaction:

2NO₂(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

1. Calculate the sum of ΔH_f⁰ for products:

ΔH_f⁰ (products) = ΔH_f⁰ (N₂O) + 3 × ΔH_f⁰ (CO₂)

ΔH_f⁰ (products) = 819 kJ/mol + 3 × (-393 kJ/mol)

ΔH_f⁰ (products) = -360 kJ/mol

2. Calculate the sum of ΔH_f⁰ for reactants:

ΔH_f⁰ (reactants) = 2 × ΔH_f⁰ (NO₂) + 3 × ΔH_f⁰ (CO)

ΔH_f⁰ (reactants) = 2 × 34 kJ/mol + 3 × (-110 kJ/mol)

ΔH_f⁰ (reactants) = -262 kJ/mol

3. Calculate the standard enthalpy change (ΔH_reaction):

ΔH_reaction = ΔH_f⁰ (products) - ΔH_f⁰ (reactants)

ΔH_reaction = -360 kJ/mol - (-262 kJ/mol)

ΔH_reaction = -98 kJ/mol

Conclusion:

The enthalpy change (ΔH) for the given reaction is -98 kJ/mol. However, since this does not match the provided options, the correct answer is: None of these

Correct answer: 2)

Concept:

• The Gibbs free energy of a system at any moment in time is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system.
• Enthalpy is the amount of heat energy transferred (heat absorbed or emitted) in a chemical process under constant pressure.
• Entropy measures the amount of heat dispersed or transferred during a chemical process.
• Reactions are classified as either exothermic (ΔH < 0) or endothermic (ΔH > 0) on the basis of whether they give off or absorb heat.
• Reactions can also be classified as exergonic (ΔG < 0) or endergonic (ΔG > 0) on the basis of whether the free energy of the system decreases or increases during the reaction.

Explanation:

• Ozone is less stable than oxygen and ozone decompose forming oxygen readily.
• 2O3 → 3O2
• The reactant ( ozone) is less stable than the product, so ΔH is negative. 
• Now, ΔS =+ve as the number of moles of gas are increasing (from 2 to 3)
• We know that  ΔG = ΔH - TΔS
• ΔG =−ve - (+ve)
• ΔG =−ve
• So, ΔG will be negative for this reaction.

Conclusion:

Thus, ΔH, ΔS and ΔG are -ve, +ve, -ve respectively