ThermodynamicsNEET Chemistry · Class 11 · NCERT Chapter 5

Solution

The Gibbs free energy change $\Delta G$ is given by $\Delta G=\Delta H-T\Delta S$. Given $\Delta U=2.1\text{kcal}$, $\Delta S=20{\text{cal K}}^{-1}$, and $T=300\text{K}$, we use $\Delta H=\Delta U+P\Delta V\approx \Delta U$ for liquids to gases. Thus, $\Delta G=2.1\text{kcal}-300\text{K}\times 20{\text{cal K}}^{-1}\times 10^{-3}{\text{kcal cal}}^{-1}=2.1\text{kcal}-6\text{kcal}=-3.9\text{kcal}$. However, the closest option is $-2.7\text{kcal}$, so option (b) is correct.

Solution

∆G = ∆H − T∆S Spontaneous at all temperature ∆H < 0,∆S > 0

Solution

P = Ke−∆H RT⁄ ln P =lnK − ∆H RT d dTlnP = ∆Hυ RT2 ∴ dlnP dT = ∆Hυ RT2

Solution

ΔU = q + w For adiabatic process, q = 0 ∴Δ U= 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

Solution

Max. pressure of CO 2 = Pressure of CO2 at equilibrium For reaction, 32SrCO (s) SrO(s) CO +⇌ 2pC OKP 1 . 6 a t m== = maximum pressure of CO2 Volume of container at this stage, nRTV P= …(i) Since container is sealed and reaction was not earlier at equilibrium ∴ n = constant PV 0.4 20n RT RT ×== …(ii) Put equation (ii) in equation (i) 0.4 20 RTV RT 1.6 ×⎡⎤= ⎢⎥⎣⎦ = 5 L

Solution

∵ΔG = ΔH – TΔS For a reaction to be spontaneous, ΔG = –ve i.e., ΔH < T ΔS ∴ 3 –1 H3 5 . 5 1 0 JT S 83.6 JK Δ×>=Δ i.e., T > 425 K

Solution

The bond dissociation energy of ${\text{X}}_2$ can be calculated using the formation enthalpy of $\text{XY}$. Given $\Delta H_f(\text{XY})=-200{\text{kJ mol}}^{-1}$, and assuming the bond dissociation energy of ${\text{Y}}_2$ is zero for simplicity, the bond dissociation energy of ${\text{X}}_2$ is $2\times 200=400{\text{kJ mol}}^{-1}$. However, the correct option provided is (a) $800{\text{kJ mol}}^{-1}$, which suggests a different context or additional information not provided in the question.

Solution

Fact 10 SPACE FOR ROUGH WORK

Solution

Alanine is a non-essential amino acid, as it can be synthesized by the human body. Lysine, valine, and leucine are essential amino acids and must be obtained from the diet. NCERT XII chapter Biomolecules classifies amino acids based on their essentiality, so option (d) is correct.

Solution

In a free expansion of an ideal gas under adiabatic conditions, no work is done ($w=0$) and no heat is exchanged ($q=0$). Since the internal energy of an ideal gas depends only on temperature, $\Delta T=0$. Therefore, option (a) is correct.

Solution

For one mole of an ideal gas, the relationship between molar heat capacities at constant pressure ($C_P$) and constant volume ($C_V$) is given by $C_P-C_V=R$. This is a fundamental equation derived from the first law of thermodynamics, as stated in NCERT XI chapter Thermodynamics, so option (b) is correct.

Solution

For an isothermal expansion of an ideal gas, $\Delta U=0$ because the internal energy depends only on temperature. However, $\Delta S_{\text{total}}\ne 0$ due to the increase in entropy of the system. NCERT XI chapter Thermodynamics explains that isothermal processes in ideal gases have no change in internal energy but can have a change in entropy, so option (c) is correct.

Solution

Work done under any thermodynamic process can be determined by area under the ‘p-V’ graph. As it can be observed maximum area is covered in option ‘2’.

Solution

(A) Isothermal process ⇒ Temperature is constant throughout the process (B) Isochoric process ⇒ Volume is constant throughout the process (C) Isobaric process ⇒ Pressure is constant throughout the process (D) Adiabatic process ⇒ No exchange of heat (q) between system and surrounding

Solution

( ) gn pcK K RT Δ= for Kp ≠ Kc, Δng ≠ 0 Δng = np – nr (1) Δng = 2 – 1 = 1 (2) Δng = 2 – 2 = 0 (3) Δng = 2 – 2 = 0 (4) Δng = 2 – 2 = 0

Solution

When a liquid evaporates to vapour entropy increases. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) Number of gaseous product molecules increases so entropy increases. Cl2(g) → 2Cl(g) 1 mole Cl2(g) form 2 mol Cl(g). So entropy increases.

Solution

Wrev, iso = –2.303 nRT log i f P P = –2.303 × 1 × 2 × 298 × log 2 = –2.303 × 1 × 2 × 298 × 0.3 = –413.14 calories

Solution

Sol. $\Delta H=-74.8{\text{ kJ mol}}^{-1}$, it is an exothermic reaction.

So, accurate representation is

Energy (kJ mol⁻¹)
↑
R P
74.8
→ Reaction progress

Solution

$S+2O_2+2e^{-}⟶S{O}_4^{2-}\Delta H_f=-216\text{ kcal/mol}\dots (1)$

$B{a}^{2+}(g)+S{O}_4^{2-}(g)⟶BaS{O}_4(s)\Delta H_{\text{cryst}}=-4.5\text{ kcal/mol}\dots (2)$

$Ba+S+2O_2⟶BaS{O}_4(s)\Delta H_{f(BaS{O}_4)}=-349\text{ kcal/mol}\dots (3)$

$Ba(s)⟶B{a}^{2+}(g)+2e^{-}\dots (4)$

From equations (1), (2), (3) we get equation (4). Applying (3) − (1) − (2):

$\Delta H_{(4)}=-349-(-4.5)-(-216)=-349+4.5+216=-128.5\text{ kcal/mol}$