NCERT Class 11 Chemistry Chemical Thermodynamics Long Question And Answers

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NCERT Class 11 Chemistry Chemical Thermodynamics Long Question And Answers

Question 1. Classify the following systems into open, closed, or isolated:

  1. Living cell,
  2. A gas is enclosed in a cylinder fitted with a movable piston.
  3. The walls ofthe container and the piston are impermeable and thermally insulated.
  4. The substances present in a soda water bottle. The chemicals participated in a chemical reaction occurring in a closed glass container.
  5. Hot tea is kept in a thermos flask.

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Answer:

  1. An open system: A living cell exchanges matter and energy with its surroundings to maintain itself.
  2. A closed system: The walls of the container and the piston are impermeable and thermally insulated. So, the system cannot exchange matter and heat with its surroundings.  However, as the piston is movable, the system can do work or work can be done on it if the pressure on the die piston is decreased or increased. So, the system can exchange energy in the form of work with its surroundings,
  3. A closed system: Here, the components present in the bottle constitute the system. As the bottle is closed, the system cannot exchange matter with its surroundings. However, it can exchange heat (energy) with its surroundings.
  4. A closed system: Here, the chemicals constitute the system. As the reaction container is closed, the system is unable to exchange matter with its surroundings. However, it can exchange heat (energy) with its surroundings,
  5. An isolated system: The walls of a thermos flask are made up of insulating materials. Again, the mouth of the flask is closed. So, the system can exchange neither matter nor energy with its surroundings.

Question 2. Identify the following an extensive or intensive property: Enthalpy, internal energy, pressure, viscosity, heat capacity, density, electric potential, specific heat capacity, molar volume, surface tension, universal gas constant, vapour pressure, number of moles, refractive index, entropy.
Answer:

An extensive property ofa system depends upon the mass ofthe substance present in the system. Its value increases as the amount of substance in the system increases. Enthalpy, internal energy, heat capacity, number of moles, and entropy are extensive properties.

An intensive property of a system is independent of the amount ofthe substance present in the system. Pressure, viscosity, density, electric potential, specific heat capacity, molar volume, surface tension, universal gas constant, vapour pressure, and refractive index are intensive properties

Question 3. Thermodynamic state functions are path-independent quantities. Explain with an example.
Answer:

A state function is a path-independent quantity. This means that when a system undergoes a process, the change in any state function depends only on the initial and final states of the system, and not on the path of the process.

To make it clear, let us consider the following process in which the mol of an ideal gas changes its state:

2 atm, 4L, 273K→1 atm, 8L, 273K

We can make this change by either of the following two processes. However, in each of these processes, the change in a state function, viz., P or V is the same.

Question 4. Why is the change in any state function in a cyclic process zero? Is the change in any function both reversible and irreversible cyclic processes zero?
Answer:

  1. The value of any state function of a system depends only on the present state of the system. In a cyclic process, the initial and the final states of a system are the same, and so are the values ofa state function at these two states.
  2. Hence, the change in a state function will be zero in a cyclic process. The change in any state function depends only on the initial and the final states ofthe system.
  3. It does not depend on the path followed to carry out the change. This means that a state function undergoes the same change in a process with the specified initial and final states irrespective of whether the process is carried out reversibly or irreversibly.
  4. Now, in a cyclic process, the change in any state function is always zero. Hence, for both reversible and irreversible cyclic processes, the change in any state function will be zero.

Question 5. 1mol of an ideal gas participates in the process as described in the figure.

  1. What type is the overall process?
  2. Is this an isothermal process?
  3. Mention the isobaric and isochoric steps in this process.

Answer:

  1. It is a cyclic process because the system returns to its initial state after performing a set of consecutive processes.
  2. This is not an isothermal process, because the temperature of the system does not remain constant throughout the process although the initial and the final temperatures are the same.
  3. Step BC represents an isobaric process because the pressure ofthe system remains constant in this step. Step CA represents an isochoric process because the volume ofthe system remains constant in this step.

Question 6. Are the following changes reversible or irreversible? Give proper explanations:

  1. Melting of ice at 0°C and 1 atm pressure
  2. The pressure ofa gas enclosed in a cylinder fitted with a piston is 5 atm.
  3. The gas is expanded against an external pressure of 1atm.

Answer:

  1. At 0°C and 1 atm pressure ice remains in equilibrium with water. If the temperature ofthe system is increased by an infinitesimal amount, ice melts into the water slowly. Again, if the temperature of the system is decreased by an infinitesimal amount, water freezes into ice slowly.
  2. Thus, an infinitesimal increase or decrease in temperature causes a change in the direction of the process. Hence, the melting of ice at 0°C and 1 atm pressure can be considered as a reversible process,
  3. An irreversible process: The external pressure is considerably less than the pressure of the gas. So, the gas will expand rapidly without maintaining thermodynamic equilibrium during the process. Hence, this expansion will occur irreversibly.

Question 7. In the process A→ B→ C, the change in internal energy of the system in the steps A→ B and B→ C are -x kJ.mol¯1 and y kj-mol-1, respectively. What will be the change in the internal energy ofthe system in step C→ A?
Answer:

A→ B→ C

Given, ΔUA→ B = -x KJ. mol-1 and ΔUB→ C = -x KJ. mol-1

∴ ΔUC→A  = -ΔUA→C = (x-y) kJ. mol-1

[Since U is a state function, its change in die forward direction of a process is the same as that in the backward direction but opposite in sign].

Question 8. A certain amount of a gas participates in the cyclic process ABCD (follow figure). Calculate the total work done g in the process
Answer:

Pressure volume work, w = -Pex(V2-V1)=-PexΔV

In step: AB: Pex = x atm, ΔV= (2y-y)L=yL

∴ w1 = -Pex ΔV = -xy L . atm

In steps BC and DA. work done is zero because the system remains constant in these steps. Instep CD: Pex = 0.5x atm and AV = (y- 2y) L = -y L.

∴ ω2 = -PexΔV = —0.5x (-y L) = 0.5.xy} L . atm

So, the total work done in the process, ω = ω1 + ω2

= (—xy + 0.5xy) L .atm =-0.5xy L.atm

= -0.5xy x 101.3 J [since 1Latm = 101.3 J] =-50.65xy J

Question 9. A particular amount of gas participates separately in the two processes given below: Process-1 For which process, the work done is maximum?
Answer:

⇒ Process-1:

Step-1: ω1 = -Pex-ΔV = -P1(V2– V1)

⇒ Step-2: ω2 = 0 [Since the volume of  the system is constant]

⇒ Total work, ω = ω1 + ω2 = -P1(V2– V1)

⇒ So, |ω| = |P1(V2-V1)|

⇒ Process-2:

Step-1: ω1 = 0 [∴ the volume of the system is constant]

⇒ Step-2: ω2 = —Pex ΔV = -P2(V2-V1)

⇒ Total work, ω’ = ω1+ ω2 = -P2(V2– V1)

⇒ So, |ω’| = |P2(V2-V1)| [Since P2 < P1 , |ω’| < |ω’| ]

Question 10. For an ideal gas, the isothermal free expansion and adiabatic free expansion are the same processes— Explain For chemical changes, why is the change in enthalpy more useful than the change in internal energy?
Answer:

According to the first law of thermodynamics, ΔU = q+ω. In the free expansion of a gas, w = 0. Again, in an adiabatic process, q = 0 Hence, in an adiabatic free expansion of an ideal gas, the change in internal energy ΔU =q + 0=0.

Also, in an isothermal process, the change in the internal energy of an ideal gas is zero. Again for free expansion of an ideal gas, w = 0. So, in an isothermal free expansion of an ideal gas, ΔU = q + w or, 0 = q+ 0 or q = 0. Therefore, it can be concluded that both processes are the same.

Question 11. The heat required to raise the temperature of 1 mol ofa gas by 1°C is q at constant volume and q’ at constant pressure. Will q be better than, less than or equal to q’? Explain
Answer:

The heat required to raise the temperature of1 mol of gas by 1 ° at constant pressure is greater than that required at constant volume. At constant pressure, the heat absorbed by a gas is used up in two ways.

One part of it is used by the gas for doing external work, and the remaining part is utilised for increasing the temperature of the gas.

At constant volume, the heat absorbed by a gas is completely utilised for increasing the temperature of the gas as no external work (P-Vwork) is possible at constant volume. Therefore, q’ must be greater than q.

Question 12. A 0.5 mol sample of H2(g) reacts with a 0.5 mol sample of Cl2(g) to form 1 mol of HCl(g). The decrease in enthalpy for the reaction is 93 kj. Draw an enthalpy diagram for this reaction.

⇒ \(\frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{g}) ; \Delta H=-93 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

The reaction is associated with a decrease in enthalpy. So, it is an exothermic reaction. In such a reaction, the total enthalpy of the product(s) (2Hp) is less than that of the reactant(s)(£flfi). Therefore, in the enthalpy diagram for the reaction, ZHp lies below ZHR.

Question 13. A 1 mol sample of N2 (g) reacts with 1 mol of O2(g) to form 2 mol of NO2(Og), where the increase in enthalpy is 180.6kj. Draw An enthalpy diagram for this reaction.
Answer:

⇒ \(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g) ; \Delta H=+180.6 \mathrm{~kJ}\)

In the reaction, the enthalpy increases. So, it is an endothermic reaction. In such a reaction the total enthalpy ofthe product(s)(ΣHR) is greater than that ofthe reactant(s) (ΣHp). Therefore, in the enthalpy diagram for the reaction, ΣHp lies above ΣHR

ΣHp

ΣH  =  +180.6 kJ

ΣHR

Question 14. Identify the exothermic and endothermic changes:
Answer: 

⇒ \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{NO}_2(g)+57.0 \mathrm{~kJ}\)

Answer:

⇒  \(\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{NO}_2(\mathrm{~g})+57.0 \mathrm{~kJ}\).

Heat is realed in this reacton,. so it’s an exothermic reaction.

H2O(s) + 6.02 kJ→H2O(l) . Heat is absorbed in this reaction. So, it is an endothermic process.

⇒ \(\mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+\mathrm{H}_2(g)-130 \mathrm{~kJ}\)

⇒ \(\text { i.e. } \mathrm{C}(s)+\mathrm{H}_2 \mathrm{O}(g)+130 \mathrm{~kJ} \rightarrow \mathrm{CO}(g)+\mathrm{H}_2(g)\)

Question 15. Write down the thermochemical equations for the following reactions:

  1. A 1mol sample of methane gas reacts with 2 mol of oxygen gas to form lmol of carbon dioxide and 2mol of water. In this reaction, 890.5 kl of heat is produced.
  2. A 1mol sample of carbon (graphite) reacts with 1 mol ofoxygen to form 1 mol of carbon dioxide gas. The heat evolved in this reaction is 393.5 kj.
  3. 6 mol of carbon dioxide gas reacts with 6 mol of water to form 6 mol of oxygen gas and lmol of glucose. The heat absorbed in this reaction is 2200 kj.

Answer:

⇒ \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)+890.5 \mathrm{~kJ}\)

⇒  \(\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \( \Delta H=-890.5 \mathrm{~kJ}\)

⇒ \(6 \mathrm{CO}_2(g)+6 \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(s)+6 \mathrm{O}_2(g)-2800 \mathrm{~kJ}\)

⇒ \(\text { i.e., } 6 \mathrm{CO}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{~s})+6 \mathrm{O}_2(\mathrm{~g}) \text {; }\)

⇒  \( \Delta H=-2800 \mathrm{~kJ}\)

Question 16. \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l); \dot{2} \Delta \mathrm{H}=-285.8 \mathrm{~kJ}\)  What will be the value of ΔH for the reaction: 2H2O(l)→ 2H2(g) + O2(g)?
Answer:

⇒  \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) \dot{2} \Delta H=-285.8 \mathrm{~kJ}\)

If this equation is written in the reverse manner, we have

⇒ \(\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) ; \Delta H=+285.8 \mathrm{~kJ}\)

Multiplying this equation by 2, we have

⇒ \(2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{H}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) ; \Delta H=4571.6 \mathrm{~kJ}\)

Question 17. Why is ΔH = ΔU for the following two reactions? Explain.

  1. NaOH(aq) + HCl(aq)→ NaCl(aq) + H2O(l)
  2. CHΔ(g) + 2O2(g) -> CO2(g) + 2H2O(g)

Answer: This reaction occurs In a solution. For a reaction occurring in a solution, All = AH

For this reaction An (total number of moles of gaseous products – total number of moles of gaseous reactants) = (1 +2)-(l + 2) = 0. So, Δ7 = ΔU according to the relation ΔH = ΔU + ΔT.

Question 18. Give an example of a reaction for each of the following relations between AH and ΔH: 

  1. ΔH<ΔH
  2. ΔH > ΔH
  3. ΔH = AH.

Answer:

In a reaction, if a gaseous substance (either as a reactant or as a product or both) participates, the change in enthalpy in the reaction at constant pressure and temperature is given by ΔH= AU + ΔnRT. K An (total number of moles of gaseous products – total number of moles of gaseous reactants) >0, < 0 or =0, then AH > AH, AH < AH or All = AH respectively.

⇒ \( 2 \mathrm{H}_2 \mathrm{O}(g) \rightarrow 2 \mathrm{H}_2(g)+\mathrm{O}_2(\mathrm{~g})\)

⇒ \(\Delta n=(2+1)-2=+1 \text {. So, } \Delta H>\Delta U\)

⇒ \(2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)\)

⇒  \(\Delta n=0-(2+1)=-3 \text {. So, } \Delta H<\Delta U\)

⇒ \(\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}(g)\)

⇒ \(\Delta n=2-(1+1)=0 . \text { So, } \Delta H=\Delta U\)

Question 19. In which of the following reactions At 25°C, does the standard enthalpy change correspond to the standard enthalpy of formation of H2O(Z)? Give reasons.

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{3}{2} \mathrm{O}_3(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)\(2 \mathrm{H}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(l)\)
Answer:

  1. The standard enthalpy change in this reaction does not indicate the standard heat of formation of H2O(Z) because, the standard state of oxygen at 25 °C, is O2(g) and not O3(g).
  2. The standard enthalpy change in this reaction refers to the standard heat of formation of H2O(Z) because 1 mol of H2O(Z) is formed from its stable constituent elements.
  3. The standard enthalpy change in this reaction does not indicate the standard heat of formation of H2O(Z). This is because the stable state of hydrogen at 25 °C, is not H(g).

Question 20. Which one of the given reactions indicates the formation reaction ofthe compound produced in the reaction?

  1. S (monoclinic) + O3(g)→ SO3(g)
  2. C (graphite, s) + 2H2(g)→CH4(g)
  3. N2(g) + O2(g)→2NO(g)

Answer:

In the formation reaction of a compound, 1 mol of the compound is formed from its constituent elements. S(s, monoclinic) +O2(g)→SO2(g), at 25 °C this reaction does not represent the formation of S02(g) because, the stable form of sulphur is S(s, rhombic) at 25°C.

2. C(s, graphite) + 2H2(g)→CH4(g), at 25 °C reaction represents the formation reaction of CH4(g) because lmol of CH4(g) is formed from its stable constituent elements.

3.  N2(g) + O2(g)→2NO(g), at 25°C this reaction does not represent the formation reaction of NO(g) because 2 mol of NO(g) are formed in the reaction.

Question 21. The standard heats of combustion of CH4(g) and C2H6(g) are -890 kj.mol-1 and -1560 kj. mol-1 respectively. Why is the calorific value of C2H6(g) lower than that of CH4 (g)?
Answer:

The enthalpy of combustion of a compound is always negative. So, the standard enthalpy of combustion of the given compound, ΔHºC = -Q kj .mol-1. The thermochemical equation for this combustion reaction

⇒ \(\mathrm{C}_x \mathrm{H}_y(l)+\left(x+\frac{y}{4}\right) \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)+\frac{y_2}{2} \mathrm{H}_2 \mathrm{O}(l)\)

⇒ \(\Delta H_c^0=-Q \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Question 22. Identify whether the enthalpy of the initial state is greater than, less than or equal to that of the final state in the following changes: solid;→ liquid; vapour→liquid →vapour
Answer: Solid→Liquid:

It is an endothermic process. In this process

⇒ \(\Delta H>0 \text {, i.e., } H_{\text {liquid }}-H_{\text {solid }}>0 \text { or, } H_{\text {liquid }}>H_{\text {solid }} \text {. }\)

Therefore, the enthalpy of the final state will be greater than that ofthe initial state.

Vapour→Solid: It is an exothermic process. So, in this process

Question 23. What docs Mi signify in each of the following equations?

  1. HCl(g) + 5H2O(l)→HCl(5H2O); ΔH = -64 kJ
  2. HCl(g) + aq→HCl(aq); ΔH = -75 kJ
  3. HCl(5H2O)+20H2O(l)→HCl(25H2O); ΔH=-8.1 kJ

Answer:

In this process, 1 mol of HCl dissolves in 5 mol of water, forming a solution of definite concentration. The enthalpy change in such a process is known as the integral heat of the solution. So, ΔH, in the process Indicates the integral heat of the solution.

In this process, 1 mol HCl dissolves in a large amount of water, forming an infinitely dilute solution. The enthalpy change in such a process is called the heat of solution. So, vH, in process 2, indicates the heat of the solution.

This is a dilution process because a solution with a definite concentration is diluted by adding solvent to it. So, ΔH, in this process, indicates the heat of dilution.

Change in enthalpy remains the same whether a reaction is carried out in one step or several steps under similar reaction conditions Explain the rearms.

Question 24. Given (at 25°C and 1 atm pressure):

  1. C (s, diamond) + O2(g)→CO2(g); ΔH°=-393.5 kj-mol-1
  2. C (s, graphite) +O2(g)→CO2(g); ΔH°=-391.6 kj-mol-1

Find the standard heat of transition from graphite to diamond.
Answer:

⇒ \(\mathrm{C}(s, \text { diamond })+\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒  \(\Delta H^0=-393.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

⇒ \(\mathrm{C}(s, \text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})\)

⇒  \(\Delta H^0=-391.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

[Subtracting equation 1 From equation We get C(s, graphite)→ C(s, diamond); AH° = 1.9 kj. mol-1 So, the heat of transition for this process is 1.9 kj.mol-1

Question 25. At 25°C, if standard enthalpies of formation o/MX(s), M+(ag) and X-(aq) are -x, y and -zkj-mol-1 respectively, then what will the heat of reaction before the reaction M+(aq) + X~(aq)-+ MX(s).
Answer:

M+(aq) + X-(aq)→MX(s)

The standard heat of reaction,

⇒ \(\Delta H^0=\Delta H_f^0[\mathrm{MX}(s)]-\Delta H_f^0\left[\mathrm{M}^{+}(a q)\right]-\Delta H_f^0\left[\mathrm{X}^{-}(a q)\right]\)

=\((-x-y+z) \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

Question 26. At 25°C, the bond dissociation energy of N2(g) is 946 kj.mol-1. What does it mean? What would be the standard atomisation enthalpy of N2(g) at 25°C?
Answer:

At 25°C, the standard bond dissociation energy of N2(g) is 946 kj.mol-1. This means that the energy required to break 1 mol of N=N bonds completely in the gaseous state to form gaseous nitrogen atoms is 948 kj. At 25 °C, the standard state of nitrogen is N2(g). Now, the formation of 1 mol of N(g) takes place by the following process \(\frac{1}{2} \mathrm{~N}_2(g) \rightarrow \mathrm{N}(g)\)

Since 1 mol of N(g) is produced from N2(g) in the process [1], the enthalpy change in this process at 25 will be equal to the standard atomisation enthalpy of nitrogen.

In process [1], change in enthalpy \(=\frac{1}{2}\) x bond dissociation energy of N=N

= \(\frac{1}{2} \times\) 946 = 473 k.mol-1 Therefore, at 25 °C the standard atomisation enthalpy of nitrogen is 473 kJ.mol-1.

Question 27. A—B bonds present in AB3(g) molecule undergo stepwise dissociation by the following sequence of steps.

  1. AB3(g)→AB2(g) + B(g) 
  2. AB2(g)→AB(g) + B(g)
  3. AB(g)→A(g) + B(g)

Answer:

Bond energy of A — B bond in AB3(g) molecule = the average bond dissociation energy of three A— B bonds in AB3(g) molecule. If the standard enthalpy change in step is ΔH° kj.mol-1, then the bond dissociation energy of the A-B bond in AB3(g) molecule

= \(\frac{x+\Delta H^0+z}{3} \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \text {. }\)

Now, the bond dissociation energy of the A-B bond = kJ.mol-1

⇒ \(y=\frac{x+\Delta H^0+z}{3} \quad \text { or, } \Delta H^0=3 y-(x+z) \mathrm{kJ} \cdot \mathrm{mol}^{-1}\)

Therefore, the standard enthalpy change in step (2)is [3y-(x+z)kJ.mol-1

Question 28. The following changes are performed on 1 mol of N2 gas

  1. Pressure Is decreased at a constant temperature
  2. Volume is decreased at a constant temperature
  3. What will be the sign of Ssys In These changes?

Answer:

If the pressure of a gas is decreased at a constant temperature, the volume of the gas increases. At a larger volume of a gas, the gas molecules get a greater chance to move about. As a result, the randomness of the molecules increases, which increases the entropy ofthe gas. Hence, A = +ve.

At constant temperature, the decrease in the volume of a gas reduces the availability of space for the movement of gas molecules. This results in a decrease in the randomness of the molecules; consequently, the entropy ofthe gas decreases. Hence, Δ = -ve

Question 29. The following two reactions occur spontaneously. What will be the signs of AS and ASsurr in these two reactions?

  1. A(s)→ B(s) + C(g); ΔH > 0
  2. 2X(g)→ X2(g); ΔH < 0

Answer:

At a given temperature and pressure, the change in entropy of the surroundings, \(\Delta S_{\text {surr }}=-\frac{\Delta H}{T}\)…………………(1)

In this reaction, ΔSsys = + ve is the gaseous substance produced in the reaction. Since, in this reaction, ΔH > 0; according to equation ΔSsurr = – ve.

In this reaction, Δ Ssys = -ve as the number of gas particles decreases. Again, in this process, AH < 0. So, according to the equation, ΔSsurr > 0

Question 30. For a reaction ΔH > 0, and another ΔH < 0. For both the reactions ΔSsys < 0. Which one is likely to occur spontaneously? Which one always occurs nonspontaneous? define

  1. Gibbs free energy
  2. The standard energy of formation of a substance
  3. The standard free energy change in a chemical reaction.

Answer:

For the reaction with ΔH>0, ΔSsurr. is -ve. Again, for this reaction ΔSsys < 0, and hence ΔSsys, + ΔSsurr < 0 i.e.  ΔSuniv < 0. So, this reaction will be non-spontaneous.

For the reaction with ΔH < 0,  ΔSsurr is -ve. Again, for this reaction ΔSsurr < 0, Therefore,

ΔSuniv = (ΔSsys, + ΔSsurr may be Positive or negative depending on the magnitudes of Ssys and Ssurr If |ΔSsys|< |ΔSsurr| then the reaction will be spontaneous.

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