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

Concept:-

• Any heterogeneous system is composed of a large number of homogeneous systems, separated by distinct boundaries. Each homogeneous system is known as a phase.
• The Gibbs phase for a multicomponent system is given by,

F = C - P + 2

where F is the degrees of freedom

C is the number of components

and P is the number of different phases.

• For one component (C=1) system with N number of phases, the number of phase lines will be,

^NC₂

Explanation:-

• For sulfur there are four phases, these are
  • Rhombic (S)
  • Monoclinic (S)
  • Liquid (L)
  • Vapour (V)

where, S→ solid, L→ liquid, V→vapour

• Thus in sulfur, there exists a total of four phases (N=4).
• Thus, the number of phase lines will be

= 4C2 

= \(\frac{4!}{\left ( 4-2 \right )!\times 2!}\)

= \(\frac{4!}{2!\times 2!}\)

= 6

Conclusion:-

• Hence, the number of phase lines observed for the phase diagram of sulfur will be 6

Concept:-

• Trouton's rule states that the entropy of vaporization (\(\ S_{vap}\)) is almost the same value, for various kinds of liquids at their boiling points. The entropy of vaporization is defined as the ratio between the enthalpy of vaporization and the boiling temperature (\(\ S_{vap}= {\frac{H_{vap}}{T_{b}}}\)).
• This rule predicts that at a normal boiling point, the molar entropy change of a liquid is constant and nearly 21.
• Mathematically Trouton's rule can be given by,

\(\Delta \bar{S}_{vap} = 21\; cal/mol\)

Explanation:-

• Trounton's rule fails in the case of associated liquids and also in the case of liquids having very high and very low boiling points. 
• Liquids having strong molecular interaction, resist molecular motion and hence show a significant deviation from Trouton's rule.
• The characteristic of those liquids to which Trouton’s rule cannot be applied is because of their special interaction between molecules, such as hydrogen bonding.
• The entropy of vaporization of formic acid is lower as predicted by Trouton's rule. This fact indicates the existence of an orderly structure in the gas phase. Formic acid exists as a dimer even in the gas phase, thus this ordered structure reduces the entropy value.
• Thus, according to Trouton's rule Large derivation from Trouton's rule is observed for systems which are having more ordered structure.

​Conclusion:-

• Hence, option 1 is correct.

Concept: 

The Clausius–Clapeyron relation characterizes a discontinuous phase transition between two phases of matter of a single constituent. On a pressure-temperature (P–T) diagram, the line separating the two phases is known as the coexistence curve. The Clausius–Clapeyron relation gives the slope of the tangents to this curve. The Clausius–Clapeyron relation can be used to find the relationship between pressure and temperature along the phase.

Explanation: 

The given phase diagram is as follows: 

We know that the melting point is the temperature at which liquid changes into solid or vice-versa.

As the diagram depicts by the dotted line between solid and liquid as the pressure increases the melting point is decreasing. 

Thus, based on the given diagram the statement: 'Melting point increases with pressure ', is wrong.

Hence, the correct option is (1).

Conclusion:-

So, the INCORRECT statement is Melting point increases with pressure

Concept:

• A phase diagram is a representation of states of a substance according to varied temperature and pressure.
• Generally, Temperature is kept on the X-axis and Pressure is kept on the y axis.
• When we cross the curves of the phase diagram, we go from one phase to another phase.
• Phase diagrams basically demonstrate the effect of temperature and pressure on the state of matter.
• At the boundary of the phase curve, the matter exists in equilibrium between two-state.
• At the triple point denoted by Tc, matter exists in all three states, solid, liquid and gas.

Explanation:

• The phase diagram of carbon dioxide is given below:
  • ​
• Extrapolating the graph, we see that:

• From the graph, we see that below Tc, it does not exist in a liquid state so, ​below Tc, it does not exist in the liquid state is a true statement.
• From the graph, we see that if we increase the temperature above the triple point, it exists in a liquid state when pressure is increased.
• So, above Tc, it does not exist in the liquid state is a false statement.
• At the critical temperature, all three phases exist, so at Tc, it can exist in all three phases is a correct statement.
• From the graph we see that if we increase the pressure after T1, solid-state does exist, so, the statement above T1, it does not exist in solid-state is a false statement.

Therefore, 1 and 2 are the correct option.

Concept:

Gibb's Free Energy:

• The decrease in free energy is actually the amount of maximum work done by the system excluding the expansion work done when temp and pressure are kept constant.
• Free energy of a system is the difference in energy at the initial state and the equilibrium state energy.
• This free energy can be used to do external work.
• The non-available energy is the equilibrium state energy and is given by T × S, where T = temperature and S = Entropy.

expressions for Gibb's Free energy are:

ΔG = ΔH - TΔS

dG = dq -dq rev

ΔGo = -nFE0​

ΔG = ΔGo + RTlnKc

• Gibbs free energy is denoted by ΔG and has the units J/mol.

Gibbs Free Energy and EMF of a cell:

• In a voltaic cell, the heat energy liberated during the chemical change is converted into electrical energy.
• The electrical energy produced for one equivalent of product is FE, where F = faradays constant 96500
• The electrical energy produced for 'n' equivalents is nFE.
• According to Helmholtz, the decrease in free energy of a reaction is equal to the electrical energy obtained from a galvanic cell. 

ΔGo = -nFE0​........................1

At Equilibrium, 

Eo =\({RT\over nF} lnK_c\).....................2, where Kc = Equilibrium constant Keq.

Equating Relation 1 and 2, we get:

\(ΔG^o = -nF​​​​{RT\over nF}lnK_{eq}\)

Or, ΔG° = - 2.303 RT Iog Keq

Hence, the relationship between the equilibrium constant K and Δ G is:​ 

ΔG° = - 2.303 RT Iog Keq

Calculation:

Given:

• ΔG° for the reaction 3 NO (g) ⇌ N2O (g) + NO2 (g) = −104.18 kJ = -104180Joules
• Temperature of the reaction = 25C = 298K
• Value of R = 8.314 J mol−1 K−1
• We know that: 

ΔG° = - 2.303 RT Iog Keq

• So substituting the given values in the above equation, we get:

-104180Joules​ = -2.303 × 8.314 J mol−1 K−1× 298 K ×  Iog Keq

or, Iog Keq = 18.255

or, K_eq = 1.8 × 10¹⁸.

Hence, the equilibrium constant for the reaction is Keq = 1.8 × 1018.

Concept:

Gibbs Free Energy (ΔG) and Heat of Reaction

ΔG = ΔH - TΔS

• Gibbs free energy (ΔG) is a thermodynamic quantity used to predict the spontaneity of a reaction at constant temperature and pressure.

Explanation:

Given:

• ΔG(25°C) = -33.089 kJ,
• ΔG(35°C) = -28.018 kJ

Therefore, the heat of the reaction is approx -184.20 kJ.

Concept:

Partial Molar Gibbs Free Energy

• The partial molar Gibbs free energy is the change in Gibbs free energy for a system when one mole of a component is added, keeping the temperature, pressure, and amounts of all other components constant.
• For an ideal gas, the partial molar Gibbs free energy, denoted by μᵢ, is expressed as the sum of the chemical potential of the pure substance (μᵢ(pure)) and a term that depends on the composition of the mixture (RT ln xᵢ), where xᵢ is the mole fraction of the component in the mixture.

Explanation:

• The given expression for partial molar Gibbs free energy in an ideal gas is:
  • μᵢ = μᵢ(pure) + RT ln xᵢ
• Here, μᵢ(pure) represents the chemical potential of the pure component, and RT ln xᵢ is the term that accounts for the non-ideal behavior of the component in the mixture.
• This expression is essential in determining the chemical potential of components in a mixture and is used in thermodynamic calculations involving ideal gases.

Therefore, the correct answer is: Option 3

Concept:

Work Done in Isothermal Reversible Expansion

• For an ideal gas expanding isothermally, the work done is calculated using the equation:

  \(W = -2.303nRT log(P_1/P_2)\)

  where:
  • W = Work done
  • n = Number of moles of gas
  • R = Universal gas constant (8.314 J/mol·K)
  • T = Temperature (in Kelvin)
  • P₁ = Initial pressure
  • P₂ = Final pressure

Explanation:

Given:

• n = 1 mole
• T = 300 K
• P₁ = 10 atm
• P₂ = 1 atm
• R = 8.314 J/mol·K

Using the formula for work done:

\(W = -2.303nRT log(P_1/P_2)\)
\(W = -2.303 \times1 \times 8.314 \times 300 \times log(10/1)\\ W = -2.303 \times1 \times 8.314 \times 300 \times log(10)\\ W = -2.303 \times1 \times 8.314 \times 300 \times 1\\ W ≈ 5744\ J\\ W ≈ 5.7 \ kJ\\\)

Therefore, the work done during the isothermal reversible expansion is approximately 5.7 kJ.

Concept:

In a chemical equilibrium, the molar composition of a mixture of substances can be described in terms of their chemical potentials. For an equilibrium mixture of components P and Q, any small disturbance in composition will affect the chemical potentials of these components. The ratio of the changes in chemical potentials can be related to the stoichiometric coefficients of the reaction involved.

• Chemical Potential: The chemical potential μ of a substance is the partial molar Gibbs free energy and an indicator of the tendency of the substance to undergo a phase change or a chemical reaction.
• Equilibrium Condition: For a reaction at equilibrium, the sum of the changes in chemical potentials, weighted by their stoichiometric coefficients, is zero:
  • ​ \(\sum x_i \delta μ_i = 0 \) where x_i are the mole fraction and \( \delta μ_i \) are the changes in chemical potentials.
• Ratio: The ratio of the stoichiometric coefficients can be determined by the ratio of the changes in chemical potentials.
  • In this case, if the chemical potential of Q changes by 15 kJ/mol and the chemical potential of P changes by -25 kJ/mol, the ratio can be expressed as:
    • ​ \( \frac{x_P}{x_Q} = \frac{\delta μ_Q}{-\delta μ_P} \)

Explanation of Statements:

Given the changes in chemical potentials, the correct ratio of stoichiometric coefficients (and hence the molar composition) is: \(\frac{x_P}{x_Q} = \frac{15}{25} = \frac{3}{5} \) Thus, the molar composition x:y (P:Q) is 3:5.

Conclusion:

The correct molar composition (x:y) of the mixture of P and Q at equilibrium is 3:5.

Concept:

For the reaction A(g) ⇌ B(g) at a particular temperature with an equilibrium constant greater than one, it indicates that the formation of product B is favored over reactant A at equilibrium. The relationship between the Gibbs free energy change (ΔG) and the equilibrium constant (K) is given by:

\(\Delta G = -RT \ln K \)

• Equilibrium Constant (K): Since ( K > 1 ), it implies that ΔG is negative, suggesting that the reaction is spontaneous in the forward direction.

• Energy Levels: The energy level diagrams for molecules A and B represent their respective stability. A lower energy level indicates a more stable state, which correlates with a favorable reaction forming B.

Explanation: 

• In the energy level diagram, molecule A is at a higher energy state compared to molecule B. This indicates that the conversion from A to B is exothermic, leading to a decrease in energy when moving to the product state.

• The transition from A to B represents a favorable energy change, which aligns with the observed equilibrium constant (K > 1). Thus, the schematic energy levels suggest that products are more stable than the reactants at equilibrium.

• As indicated in the diagram, the energy levels reflect the tendency of the system to favor products B over A, which supports the conclusion that the energy of A is higher than that of B.

Conclusion:

The correct energy level configuration, where B is more stable than A (reflecting K_eq > 1 ), is shown in Option 1.