Q: The correct option for free expansion of an ideal gas under adiabatic condition is:

Free expansion ⇒Pex=0 ∴ w=−Pex△V=0 also, △ U = q + w [ first law of thermodynamics]∴U=0 ∵ Internal energy of an ideal gas is a function of temperature∴ If internal energy remains constant∴ △T=0

Q: For the reaction, 2Cl(g)→Cl2(g) , the correct option is :

Given reaction, 2Cl(g)→Cl2(g) We know that,Cl2(g)→2Cl(g) is endothermic reaction because it requires energy to break bond.Also, two gaseous atom combine together to form 1 gaseous molecule.So, randomness △rS<0

Q: Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is :(Given that 1 L bar = 100 J)

∴Wirr=−Pext△V = – 2 bar × (0.25 – 0.1) L = – 2 × 0.15 L-bar = – 0.30 L-bar = – 0.30 × 100 J = – 30 J

Q: In which case change in entropy is negative?

• H2O(l)⇋H2O(v),△S>0 • Expansion of gas at constant temperature, △ S > 0• Sublimation of solid to gas, △ S > 0• 2H(g)→H2(g),△S<0(∵△ng<0)

Q: The correct thermodynamic conditions for the spontaneous reaction at all temperatures is:

ΔG = ΔH − TΔSSpontaneous at all temperature ΔH 0

Q: Using the Gibbs energy change,ΔG∘=+63.3 kJ , for the following reaction,Ag2CO3⇌2Ag+(aq)+CO32−(aq) the Ksp of Ag2CO3 (s) in water at 25∘C is : (R=8.314 J K−1 mol−1)

ΔG∘=−2.303RTlogKsp 63.3×1000=−2.303×8.314×298logKsp logKsp=−11.09 Ksp=10−11.09=8×10−12

Q: Which of the following statements is correct for the spontaneous adsorption of a gas?

During adsorption entropy decreases, so ΔS<0. ΔG=ΔH−TΔS For spontaneous adsorption ΔG<0 so ΔH should be highly negative.

Q: For the reaction :X2O4(ℓ)⟶2XO2( g) ΔU=2.1kcal,ΔS=20calK−1 at 300 K Hence ΔG is:

According to Le-Chatelier's Principle→ In exothermic reactions low temperature favours the forward reaction→ On increasing pressure equilibrium shifts towards less number of moles.

Q: A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5atm from an initial volume of 2.50 L to a final volume of 4.50 L . The change in internal energy ΔU of the gas in joules will be:

ΔU=q+w For adiabatic process, q=0∴ΔU=w=−P⋅ΔV=−2.5 atm×(4.5−2.5)L=−2.5×2 L−atm=−5×101.3 J=−506.5 J=−505 J

Q: For a given reaction, ΔH=35.5 kJ mol−1 and ΔS=83.6JK−1 mol−1. The reaction is spontaneous at : (Assume thatΔH and ΔS do not vary with temperature)

∵ΔG=ΔH−TΔS For a reaction to be spontaneous, ΔG=−ve i.e., ΔH ΔSΔH=83.6JK−135.5×103 J i.e., T>425 K

Question 1

The correct option for free expansion of an ideal gas under adiabatic condition is:

Accepted Answer

Free expansion ⇒Pex​=0 ∴ w=−Pex​△V=0 also, △ U = q + w [ first law of thermodynamics]∴U=0 ∵  Internal energy of an ideal gas is a function of temperature∴  If internal energy remains constant∴ △T=0

Question 2

For the reaction, 2Cl(g)→Cl2​(g) , the correct option is :

Accepted Answer

Given reaction, 2Cl(g)→Cl2​(g) We know that,Cl2​(g)→2Cl(g) is endothermic reaction because it requires energy to break bond.Also, two gaseous atom combine together to form 1 gaseous molecule.So, randomness △r​S<0

Question 3

Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is :(Given that 1 L bar = 100 J)

Accepted Answer

∴Wirr​=−Pext​△V 				= – 2 bar × (0.25 – 0.1) L				= – 2 × 0.15 L-bar				= – 0.30 L-bar				= – 0.30 × 100 J				= – 30 J

Question 4

In which case change in entropy is negative?

Accepted Answer

• H2O(l)⇋H2O(v),△S>0 • Expansion of gas at constant temperature, △ S > 0• Sublimation of solid to gas, △ S > 0• 2H(g)→H2(g),△S<0(∵△ng<0)

Question 5

The correct thermodynamic conditions for the spontaneous reaction at all temperatures is:

Accepted Answer

ΔG = ΔH − TΔSSpontaneous at all temperature ΔH  0

Question 6

Using the Gibbs energy change,ΔG∘=+63.3 kJ , for the following reaction,Ag2​CO3​⇌2Ag+(aq)+CO32−​(aq) the Ksp ​ of Ag2​CO3​  (s) in water at 25∘C is : (R=8.314 J K−1 mol−1)

Accepted Answer

ΔG∘=−2.303RTlogKsp​ 63.3×1000=−2.303×8.314×298logKsp logKsp=−11.09 Ksp=10−11.09=8×10−12

Question 7

Which of the following statements is correct for the spontaneous adsorption of a gas?

Accepted Answer

During adsorption entropy decreases, so ΔS<0. ΔG=ΔH−TΔS For spontaneous adsorption ΔG<0 so ΔH  should be highly negative.

Question 8

For the reaction :X2​O4​(ℓ)⟶2XO2​( g) ΔU=2.1kcal,ΔS=20calK−1  at 300 K Hence ΔG  is:

Accepted Answer

According to Le-Chatelier's Principle→  In exothermic reactions low temperature favours the forward reaction→  On increasing pressure equilibrium shifts towards less number of moles.

Question 9

A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5atm from an initial volume of 2.50 L  to a final volume of 4.50 L . The change in internal energy ΔU  of the gas in joules will be:

Accepted Answer

ΔU=q+w For adiabatic process, q=0∴ΔU​=w=−P⋅ΔV=−2.5 atm×(4.5−2.5)L=−2.5×2 L−atm=−5×101.3 J=−506.5 J=−505 J​

Question 10

For a given reaction, ΔH=35.5 kJ mol−1  and ΔS=83.6JK−1 mol−1. The reaction is spontaneous at : (Assume thatΔH  and ΔS  do not vary with temperature)

Accepted Answer

∵ΔG=ΔH−TΔS For a reaction to be spontaneous, ΔG=−ve  i.e., ΔH<TΔS ∴T>ΔSΔH​=83.6JK−135.5×103 J​ i.e., T>425 K

Question 11

Choose the correct answer. A thermodynamic state function is a quantity :

Accepted Answer

A thermodynamic state function is a quantity whose value is independent of path. Thus, enthalpy is a state function. Its value depends on iniital and final states and is independent of path followed.

Question 12

For the process to occur under adiabatic conditions, the correct condition is :

Accepted Answer

For the process to occur under adiabatic conditions the correct condition is q=0 .There is no exchange of heat.

Question 13

The enthalpies of all elements in their standard states are :

Accepted Answer

The enthalpies of all elements in their standard states are zero.Thus, the enthalpies ofNa(s),K(s)He(g)etc. in their standard states are zero.

Question 14

ΔUo of combustion of methane is −X kJ mole−1. The value of ΔHo is:

Accepted Answer

The balanced chemical equation for the combustion reaction is :CH4​(g)+2O2​( g)⟶CO2​(g)+2H2​O(l)Δng​=1−3=−2ΔH⊖=ΔUθ+Δng​RT=ΔUΘ−2RT ∴ΔH⊖<ΔU⊖ or (iii) is the correct answer.

Question 15

The enthalpy of combustion of methane, graphite, and dihydrogen at 298K are −890.3 kJ mol−1, 393.5 kJ mol−1  and −285.8 kJ mol−1 respectively. Enthalpy of formation of CH4​(g) will be:

Accepted Answer

According to the question,a) CH4( g)​+2O2( g)​⟶CO2( g)​+2H2​O(g)​△H=−890.3 kJ mol−1b) C(s)​+O2( g)​⟶CO2( g)​△H=−393.5 kJ mol−1c) 2H2( g)​+O2( g)​⟶2H2​O(g)​△H=−285.8 kJ mol−1Thus, the desired equation is the one that represents the formation of CH4( g)​ i.e.,C(s)​+2H2​( g)⟶CH4​( g)△f​HCH4​​=△C​HC​+2△C​HH2​​−△C​HCH4​​=[−393.5+2(−285.8)−(−890.3)]kJmol−1=−74.8 kJ mol−1∴ Enthalpy of formation of CH4( g)​=−74.8 kJ mol−1

Question 16

A reaction A+B→ C+D is found to have a positive entropy change. The reaction will be :

Accepted Answer

A reaction A+B→C+D, is found to have a positive entropy change.The reaction will be possible at any temperature.A spontaneous reactions proceed with increase in entropy.A non spontaneous reaction proceeds with decrease in entropy.

Question 17

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Accepted Answer

System absorbs 701J of heat which has positive sign and does 394J of work which has negative sign.According to the first law of thermodynamics,ΔU=q+w=701−394=307J.

Question 18

The reaction of cyanamide NH2​CN(s)  with dioxygen was carried out in a bomb calorimeter and △U was found to be −742.7kJ mol−1  at 298 K. NH2​CN(g)+23​O2​(g)→N2​(g)+CO2​(g)+H2​O(l) Calculate enthalpy change for the reaction at 298 K?

Accepted Answer

For given reaction, Δn=1+1−1.5=0.5.ΔH=ΔU+Δng​RT=−742.7+0.5×8.314×10−3×298=−742.7+1.2=−741.5 kJ mole−1

Question 19

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol−1 K−1 .

Accepted Answer

No. of moles of Al(m)=(60 g)/(27 g mol−1)=2.22 mol Molar heat capacity (C)=24 J mol−1 K−1 Rise in temperature (ΔT)=55−35=20∘C=20 K Heat evolved (q)=C×m×T=(24 Jmo∣−1 K−1)×(2.22 mol)×(20 K) =1065.6 J=1.067kj

Question 20

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0∘C to ice at −10.0∘C. △fus​H=6.03 KJ mol−1 at 0∘CCp​[H2​O(1)]=75.3Jmol−1K−1Cp​[H2​O(s)]=36.8Jmol−1K−1

Accepted Answer

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0∘C  to ice at −10.0∘C  △fus​H=6.03 KJ mol−1 at 0∘CCp​[H2​O(1)]=75.3Jmol−1K−1 Cp​[H2​O(s)]=36.8Jmol−1K−1

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