Test: Thermodynamics (August 18) - NEET MCQ

ΔS (per mole ) = 21.98 JK ^-1mol^-1

► 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.

H₂(g) + Br₂(g) → 2HBr(g)

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

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.

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).

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

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.

-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 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

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

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

∴ Correct choice : (d)

∴ Correct choice : (d)

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