NCERT Solutions for Class 11 Chemistry Part 1 Chapter 5- Thermodynamics

 Formation of CO2 from carbon and dioxygen gas can be represented as 

C(s)+O2(g) toCO2(g);ΔH=−393.5kJ,mol−1 (1mole=44g) 

Heat released in the formation of 44 g of CO2 = 393.5 kJmoll−1 

Heat released in the formation of 35.2 g of CO2=(393.5kJ)×(35.2g)(44g)=314.8kJ

So, heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas is 314.8 kJ.

 

A thermodynamic state function is a quantity whose value is independent of the path. Functions like p, V, T etc. depend only on the state of a system and not on the path. 

Hence, alternative (ii) is correct. 

 

A system is said to be under adiabatic conditions if there is zero exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q=0. Therefore, alternative (iii) is correct.

The enthalpy of all elements in their standard state is zero. Therefore, alternative (ii) is correct.

 

Since ΔHθ = ΔUθ + ΔngRT and ΔUθ=−XkJ mol−1

ΔHθ = (−X) + ΔngRT

⇒ΔHθ<ΔUθ

Therefore, alternative (iii) is correct.

 

According to the question, 

(i) CH4(g)+2O2(g)→CO2(g)+2H2O(l);ΔcHΘ=−890.3kJmol−1 

(ii) C(s)+2O2(g)→CO2(g);ΔcHΘ=−393.5kJmol−1 

(iii) 2H2(g)+O2(g)→2H2O(l);ΔcHΘ=−285.8kJmol−1 

Thus, the desired equation is the one that represents the formation of CH4(g)that is as follows:

C(s)+2H2(g)→CH4(g);ΔfHCH4=ΔcHc+2ΔcHH2−ΔcHCO2

Substituting the values in the above formula :

Enthalpy of formation CH4(g)= (−393.5)+2×(−285.8)−(−890.3)=−74.8kJmol−1 

Therefore, alternative (i) is correct.

 

For a reaction to be spontaneous, ΔG should be negative

ΔG = ΔH − TΔS 

According to the question, for the given reaction, 

ΔS = positive 

ΔH= negative (since heat is evolved) 

That results in

ΔG= negative

Therefore, the reaction is spontaneous at any temperature. 

Hence, alternative (iv) is correct. 

 

According to the first law of thermodynamics, ΔU= q + W....(i)

Where, 

ΔU = change in internal energy for a process 

q = heat 

W = work 

Given, 

q = + 701 J (Since heat is absorbed) 

W = -394 J (Since work is done by the system)

Substituting the values in expression (i), we get 

ΔU= 701 J + (−394 J)

 ΔU= 307 J

Hence, the change in internal energy for the given process is 307 J.

 

Enthalpy change for a reaction (ΔH) is given by the expression, 

ΔH = ΔU + ΔngRT

Where, 

ΔU = change in internal energy 

Δng = change in number of moles 

For the given reaction, 

Δng=∑ng (products) - ∑ng(reactants)

Δng=(2−1.5)moles

Δng=+0.5moles

And, kJ mol−1 

T = 298 K 

R = 8.314×10−3kJmol−1K−1

Substituting the values in the expression of

ΔH = (−742.7 kJ mol−1) + (+0.5 mol) (298 K)8.314×10−3kJmol−1K−1

ΔH = −741.5kJ mol−1

 

From the expression of heat (q), q = m. c. ΔT 

Where, 

c = molar heat capacity 

m = mass of substance 

ΔT = change in temperature 

Given, 

m = 60 g

c = 24J mol−1K−1

DeltaT=(55−35)∘C

ΔT=(328−308)K=20K

Substituting the values in the expression of heat:

q=(6027mol)(24Jmol−1K−1)(20K)

q = 1066.7 J

q = 1.07 kJ

 

Total enthalpy change involved in the transformation is the sum of the following changes: 

1. Energy change involved in the transformation of 1 mol of water at 10.0∘C  to 1mol of water at 0∘C.

2. Energy change involved in the transformation of 1 mol of water at 0∘Cto 1 mol of ice at 0∘C. 

3. Energy change involved in the transformation of 1 mol of ice at 0∘C to 1 mol of ice at 10∘C. 

N2O4(g)+3CO(g)→N2O(g)+3CO2(g)

 Ans: ΔrH for a reaction is defined as the difference between ΔfHvalue of products and ΔfHvalue of reactants. 

ΔrH= ∑ΔfH (product) - ∑ΔfH(reactant)

For the given reaction, N2O4(g)+3CO(g)→N2O(g)+3CO2(g)

ΔrH=[{ΔfH(NO2)+3ΔfH(CO2)}−{ΔfH(N2O)+3ΔfH(CO)}] 

Substituting the values of

ΔfHfor CO (g), CO2(g), N2O(g) and N2O4(g)from the question, we get: 

ΔrH=[{81kJmol−1+3(−393)kJmol−1}−{9.7kJmol−1+3(−110)kJmol−1}]

ΔrH=−777.7kJmol−1 

Hence, the value of ΔrH for the reaction is −777.7kJmol−1

 

Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state. 

Re-writing the given equation for 1 mole of NH3(g)is as follows:

12N2(g)+32H2(g)→2NH3(g) 

Therefore, standard enthalpy of formation of

NH3(g) = 1/2ΔrHθ 

1/2 (−92.4 kJ mol−1)

−46.2 kJ mol−1

 

The reaction that takes place during the formation of CH3OH(ℓ)can be written as: 

C(s)+2H2O(g)+12O2(g)→CH3OH(ℓ)(1)

 The reaction (1) can be obtained from the given reactions by following the algebraic calculations as: 

Equation (ii) + 2 × equation (iii) - equation (i) 

ΔfHθ[CH3OH(ℓ)]=ΔcHθ+2ΔfHθ[H2O(l)]−ΔrHθ

= (-393 - 572 + 726) kJmol−1

Therefore,

ΔfHθ[CH3OH(ℓ)]=−239 kJmol−1

 

The chemical equations implying to the given values of enthalpies are: 

1. CC4(l)→CCl4(g)ΔvapHθ=30.5kJmol−1 

2. C(s)→C(g)ΔaHθ=715.0 kJ mol−1 

3. Cl2(g)→2Cl(g)ΔaHθ=242 kJ mol−1 

4. C(g)+4Cl(g)→CCl4(g)ΔfH=−135.5 kJ mol−1 

 

Enthalpy change for the given process

CCl4(g)→C(g)+4Cl(g) can be calculated using the following algebraic calculations as: 

Equation (ii) + 2 × Equation (iii) - Equation (i) - Equation (iv) 

ΔH=ΔaHθ(C)+2ΔaHθ(Cl2)−Δvap Hθ−ΔfH

 (715.0 kJ mol−1)+2(242 kJ mol−1)−(30.5 kJ mol−1)−(−135.5 kJ mol−1)

Therefore,

ΔH=1304 kJ mol−1 

Bond enthalpy of C-Cl bond in CCl4(g) =13044kJmol−1

=326 kJ mol−1

 

ΔSwill be positive i.e., greater than zero. Since for an isolated system,ΔU=0, hence ΔS will be positive and the reaction will be spontaneous. 

 

From the expression, ΔG= ΔH−TΔS Assuming the reaction at equilibrium,

ΔTfor the reaction would be: 

T=(ΔH−ΔG)1ΔS =ΔHΔS

(ΔG = 0 at equilibrium) =400 kJ mol−10.2 kJ K−1 mol−1

T = 2000 K 

For the reaction to be spontaneous, ΔGmust be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

 

ΔH and ΔS are negative.The given reaction represents the formation of chlorine molecules from chlorine atoms. Here, bond formation is occurring. Therefore, energy is being released. Hence, ΔH is negative. Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased,

ΔS is negative for the given reaction. 

 

For the given reaction, 2A(g) + B(g) → 2D(g) Δng=2−(3) = -1 mole

Substituting the value of ΔUθ in the expression of ΔHθ=ΔUθ+ΔngRT 

=(−10.5 kJ)−(−1)(8.314×10−3 kJ K−1 mol−1)(298 K)−10.5kJ−2.48kJ

 ΔHθ=−12.98kJ

 

Substituting the values of ΔHθ and ΔSθ in the expression of ΔGθ: 

ΔGθ = ΔHθ −TΔSθ =−12.98kJ−(298K)(−44.1)J K−1

=−12.98kJ+13.14kJ

ΔGθ= + 0.16 kJ 

Since ΔGθ for the reaction is positive, the reaction will not occur spontaneously.

 

From the expression, ΔGθ = −2.303 RTlogKeq

ΔGθ for the reaction, 

=(2.303)(8.314 JK−1 mol−1)(300K)log10

 =−5744.14 Jmol−1

 =−5.744k Jmol−1

 

The positive value of ΔrH indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (O2). Hence, NO(g) is unstable. The negative value of ΔrH indicates that heat is evolved during the formation of NO2(g) from NO(g) and O2(g). The product,

NO2(g)is stabilized with minimum energy. Hence, unstable NO(g) changes to unstable NO2(g). 

 

It is given that 286 kJ mol−1 of heat is evolved on the formation of 1 mol of

H2O(l). Thus, an equal amount of heat will be absorbed by the surroundings.

qsurr=+286kJkJ mol−1

Entropy change (ΔSsurr) for the surroundings =286kJmol−1298K

Therefore, (ΔSsurr)= 959.73Jmol−1K−1.