31 Year NEET Previous Year Questions: Thermodynamics - 1 - NEET MCQ

ΔS (per mole ) = 21.98 JK ^-1mol^-1

ΔG = – PΔV = Work done
ΔV is the change in molar volume in the conversion of graphite to diamond.

Work done = – (–1.91 × 10^–3) × P × 101.3 J

∵ 1 atm = 10⁵ × 1.013 Pa
⇒ P = 9.92 x 10⁸ Pa

► The heat of formation is defined as the heat generated or observed when the compound is formed from its component elements in their standard state.

• By the definition (C) and (A) are incorrect as the starting material are compounds.
• (B) is incorrect as ozone is not the standard state for oxygen.
• (D) is correct as it satisfies the definition of the heat of formation.

ΔH = ΔE + nRT
Δn_g = 3 - (1 + 5) = -3
ΔH - ΔE = (-3 RT)

H₂(g) + Br₂(g) → 2HBr(g)

ΔHº = (B.E.)_react - (B.E.)_prod
⇒ ΔHº = (433 + 192) - (2 x 364) = 625 – 728 = – 103 kJ

ΔG = ΔH - TΔS
ΔG = -382.64 - (298 x 145.6 x 10^-3) = -339.3 kJ mol^-1

For a spontaneous process, ΔS_total is always positive.

W = – pΔV
W = -3(6 - 4) = - 6 L atm = - 6 x 101.32 J = (approx) - 608 J

As MgO is a oxide of  weak base hence  some energy is lost to break MgO (s). Hence enthalpy is less than –57.33 kJ mol^–1.

For a spontaneous reaction, ΔG  is always –ve.
Since, ΔG = ΔH – TΔS 
Therefore, ΔS = +ve, ΔH = +ve and  TΔS > ΔH.

The measure of the disorder of a system is nothing but Entropy.
For a spontaneous reaction, ΔG < 0. As per Gibbs Helmnoltz equation, ΔG = ΔH – TΔS.
Thus, ΔG is –ve, if ΔH = –ve (exothermic) and ΔS = +ve (increasing disorder).

= – 348.5 kJ

The resonance energy provides extra stability to the benzene molecule so it has to be overcome for hydrogenation to take place.
So, ΔH = – 358.5 – (–150.4) = –208.1 kJ

We know that, ΔG = ΔH – TΔS
When the reaction is in equilibrium, ΔG = 0

• If ΔG_system = 0 the system has attained equilibrium is correct.
• (d) is confusing as to when ΔG > 0, the process may be spontaneous when it is coupled with a reaction which has ΔG < 0 and total ΔG is negative.

We know that: ΔH = ΔE + PΔV
In the reactions, H₂ + Br₂ → 2HBr there is no change in volume or ΔV = 0.
So, ΔH = ΔE for this reaction.

Reaction (ii) shows the formation of H₂O, and the X₂ represents the enthalpy of formation of H₂O because as the definition suggests that the enthalpy of formation is the heat evolved or absorbed when one mole of a substance is formed from its constituent atoms.

-90 = [1/2 x 430 + 1/2 x 240] - B.E. of HCl
∴  B.E. of HCl = 215 + 120 + 90 = 425 kJ mol^–1

► A property whose value doesn’t depend on the path taken to reach that specific value is known as state functions or point functions.
► In contrast, those functions which do depend on the path from two points are known as path functions. 

Heat (q) and Work (w) are not state functions but (q + w) is a state functions.
H – TS (i.e. G) is also a state function.

Thus, II and III are not state functions so the correct answer is option (d).

The reaction given is an example of a decomposition reaction
Since, decomposition reactions are endothermic in nature, therefore, ΔH > 0.
Δn = (1+1) – 1= +1
Hence more molecules are present in products which shows more randomness i.e. ΔS > 0 (ΔS is positive).

The reaction of formation of HCl: H₂ + Cl₂ → 2HCI
⇒ Enthalpy of formation of 2HCl = - (862 - 676) = -186 kJ.
∴ Enthalpy of formation of HCl = -(186/2) kJ = -93 kJ

ΔG = ΔH – T Δ S
At equilibrium, ΔG = 0
⇒ 0 = (170 × 103 J) – T (170 JK^{– 1})
⇒ T = 1000 K
For spontaneity, ΔG is – ve, which is possible only if T > 1000 K.

Enthalpy of reaction = B.E_(Reactant)– B.E_(Product)
ΔH_reaction = [606.1 + (4 × 410.5) + 431.37)] –  [336.49 + (6 × 410.5)]  = –120.0 kJ mol^–1

ΔS for the reaction 

ΔS = 50 – (30 + 60) = – 40 J
For equilibrium ΔG = 0 = ΔH – TΔS

ΔH = 40630J mol ^–1
ΔS = 108.8JK^–1 mol ^–1

∴ Correct choice : (d)

∴ Correct choice : (d)

ΔH = –26.8 + 33.0 = + 6.2 kJ

∴   Correct choice : (a)

Given ΔH = 30 kJ mol^–1 T = 273 + 27 = 300 K
= 100J mol^–1 K^–1

Given:
4H_(g) → 2H_{2 (g)}; ΔH = - 869.6 kJ
i.e. 2H_{2 (g)} → 4H_(g); ΔH = 869.6 kJ

⇒ H_{2 (g)} → 2H_(g); ΔH = (869.6) * 1/2 = 434.8 kJ

To calculate ΔH operate 2 × eq. (1) + eq. (2) – eq. (3)
ΔH = 300 – 125 – 350 = – 175KJ/mol

Since in the first reaction gaseous products are forming from solid carbon hence entropy will increase i.e. ΔSº = +ve.

Since, ΔG° = ΔH° – TΔS
Hence the value of ΔG decrease on increasing temperature.