31 Year NEET Previous Questions: Thermodynamics - 2 Free MCQ Practice Test

As  ΔH = ΔE + Δn_gRT if n p < nr; Δ n_g = n_p – n_r = – v_e.
Hence Δ H < Δ E.

1MH₂SO₄ = 2g eq. of H₂SO₄.
Hence y = 2x or 

Δn_g = 2 – 4 = – 2, ΔH = ΔE – 2RT.

During isothermal expansion of ideal gas,
ΔT = 0.   Now H = E + PV
∵ ΔH = ΔE + Δ(PV)
∴ ΔH = ΔE + Δ(nRTT));
Thus if ΔT = 0., ΔH = ΔE i.e., remain unaffected

ΔG is negative for a spontaneous process.

This is combustion reaction, which is always exothermic hence
ΔH = –ve
As the no. of gaseous molecules are increasing hence entropy increases now ΔG = ΔH - TΔS
For a spontaneous reaction ΔG = -ve
Which is possible in this case as ΔH = –ve and ΔS = +ve.

If more tran s-2-pentene is added, then its concentration in right hand side will increase. But in order to maintain the K constant, concentration of cis-2-pentene will also increase. Therefore more cis-2-pentene will be formed.

Enthalpy of formation of C₂H₄ , CO₂ and H₂O are 52 , – 394 and – 2 86 k J/ mol respectively. (Given) The reaction is
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O.
change in enthalpy, (ΔH) = ΔH_products - ΔH_reactants
= 2 x (-394) + 2 x (-286) - (52 + 0)
= – 1412 kJ/ mol.

Enthalpy of formation of C₂H₄ , CO₂ and H₂O are 52 , – 394 and – 2 86 k J/ mol respectively. (Given) The reaction is
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O.
change in enthalpy, (ΔH) = ΔH_products - ΔH_reactants
= 2 x (-394) + 2 x (-286) - (52 + 0)
= – 1412 kJ/ mol.

The relation between free energy change and equilibrium constant is given by Nernst equation

At equilibrium,  E_c = 0 and Q = K_C

................(i)

Again ΔGº = nFEº (ii)

put in (i)

Entropy states the randomness or disorderness of the system. At absolute zero, the movement of molecules of the system or randomenss of the system is zero, hence entropy is also zero.

Entropy change ΔS =  ΔS_product - ΔS_reactant  
= 2 (186.7) – (223 + 130.6)
 = 373.4 – 353.6
 = 19.8 JK^–1 mol^–1

Hydration energy of Cl⁺ is very less than H⁺ hence it doesn’t form Cl⁺ (aq) ion.

Given C + O₂ = CO₂ , ΔHº =-x kJ    ....(1) 
2CO₂ = 2CO +O₂ ΔHº = + y kJ  … (2)
or CO₂ = CO+ 1/2O₂, ΔHº =+ y/2 kJ ...(3)
From eq. no. (1) and (3)

We know from the third law of thermodynamics, the entropy of a perfectly crystalline substance at absolute zero temperature is taken to be zero.

For an isother mal process ΔE  = 0

For a cyclic process

ΔE = 0, ΔH = 0 & ΔG = 0 .

As all depend upon final state and initial state,w doesn’t depend on path followed.

ΔE = ΔQ–W
For adiabatic expansion,ΔQ = 0
⇒  ΔE = –W
The negative sign shows decrease in Internal energy, which is equal to the work done on the system by the surroundings.

As all reactant and product are liquid 

= 9.77 J K^–1 mol^–1

For the reaction 2ZnS → 2Zn + S₂ ; ΔG1º = 293 kJ  ..........(1)
2Zn + O₂ → ZnO ; ΔG2º = –480 kJ  ..........(2)
S₂ + 2O₂ → 2 SO₂ ; ΔG3º = –544 kJ  .........(3)
ΔGº for the reaction 2ZnS + 3O₂ → 2ZnO + 2SO₂ can be obtained by adding eqn. (1), (2) and (3)
⇒ ΔGº = 293 – 480 – 544 = – 731 kJ

 = –98.2 + 298.2 = 200 kJ/Mole

Solution:-

As we know that, for a reaction,

ΔH=Ea​ for forward reaction −Ea​ for backward reaction.

Given that:-

Ea​ for forward reaction =19kJ

Ea​ for backward reaction =9kJ

∴ΔH=19−9=10kJ

Since the enthalpy change for the conversion of CH₄ to CH₃OH is negative, energy is released.
Therefore, CH₃OH has lower enthalpy than CH₄.

Let,

• Enthalpy of combustion of CH₄ = x
• Enthalpy of combustion of CH₃OH = y

Combustion reactions:

Since CH₄ is at a higher energy level than CH₃OH, more energy will be released when CH₄ is completely combusted compared to CH₃OH.

∴ x > y

As volume is constant hence work done in this proces is zero hence heat supplied is equal to change in internal energy.

Internal energy is dependent upon temperature and according to first law of thermodynamics total energy of an isolated system remains same, i.e., in a system of constant mass, energy can neither be created nor destroyed by any physical or chemical change but can be transformed from one form to another
ΔE = q + w For closed insulated container, q = 0, so,
ΔE = +w , as wor k is done by the system

For isothermal reversible expansion

Entropy change, 

q —→ required heat per mole
T —→ constant absolute temperature
Unit of entropy is JK^–1 mol^–1

Given C_p = 75 JK^–1 mol^–1

 Q = 1000J         ΔT = ?

Q = nC_pΔT