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QUESTION: 1

What is the entropy change (in JK^{–1} mol^{–1}) when one mole of ice is converted into water at 0º C? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol^{–1} at 0ºC) [2003]

Solution:

ΔS (per mole ) = 21.98 JK^{ -1}mol^{-1}

QUESTION: 2

The densities of graphite and diamond at 298 K are 2.25 and 3.31g cm^{–3}, respectively. If the standard free energy difference (ΔGº) is equal to 1895 J mol^{–1}, the pressure at which graphite will be transformed into diamond at 298 K is: [2003]

Solution:

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 = 10^{5} × 1.013 Pa

⇒ P = 9.92 x 10^{8} Pa

QUESTION: 3

For which one of the following equations is ΔHº_{react} equal to ΔH_{f}º for the product? [2003]

Solution:

**► **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.

QUESTION: 4

For the reaction C_{3}H_{8(g)} + 5O_{2(g) }→ 3CO_{2(g)} + 4H_{2}O_{(l) }at constant temperature, ΔH – ΔE is: [2003]

Solution:

ΔH = ΔE + nRT

Δn_{g} = 3 - (1 + 5) = -3

ΔH - ΔE = (-3 RT)

QUESTION: 5

If the bond energies of H-H, Br-Br, and HBr are 433, 192 and 364 kJ mol^{–1 }respectively, the ΔH° for the reaction H_{2}(g) + Br_{2}(g) → 2HBr(g) is: [2004]

Solution:

H_{2}(g) + Br_{2}(g) → 2HBr(g)

ΔHº = (B.E.)_{react} - (B.E.)_{prod}

⇒ ΔHº = (433 + 192) - (2 x 364) = 625 – 728 = – 103 kJ

QUESTION: 6

The standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are -382.64 kJ mol^{–1} and -145.6 JK^{–1} mol^{–1}, respectively.

Standard Gibb's energy change for the same reaction at 298 K is: [2004]

Solution:

ΔG = ΔH - TΔS

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

QUESTION: 7

Considering entropy(S) as a thermodynamic parameter, the criterion for the spontaneity of any process is: [2004]

Solution:

For a spontaneous process, ΔS_{total} is **always positive**.

QUESTION: 8

The work done during the expansion of a gas from a volume of 4 dm^{3} to 6 dm^{3} against a constant external pressure of 3 atm is: (1L atm = 101.32 J) [2004]

Solution:

W = – pΔV

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

QUESTION: 9

The absolute enthalpy of neutralisation of the reaction: MgO (s) + 2HCl (aq) —→ MgCl_{2}(aq) + H_{2}O (l) will be:[2005]

Solution:

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}.

QUESTION: 10

A reaction occurs spontaneously if: [2005]

Solution:

For a spontaneous reaction, ΔG is always –ve.

Since, ΔG = ΔH – TΔS

Therefore, ΔS = +ve, ΔH = +ve and TΔS > ΔH.

QUESTION: 11

Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction? [2005]

Solution:

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).**

QUESTION: 12

The enthalpy of hydrogenation of cyclohexene is –119.5 kJ mol^{–1}. If the resonance energy of benzene is –150.4 kJ mol^{–1}, its enthalpy of hydrogenation would be: [2006]

Solution:

= – 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

QUESTION: 13

The enthalpy and entropy change for the reaction Br_{2}(l) + Cl_{2}(g) → 2BrCl(g) are 30kJ mol^{–1} and 105 JK^{–1} mol^{–1 }respectively.The temperature at which the reaction will be in equilibrium is: [2006]

Solution:

We know that, ΔG = ΔH – TΔS

When the reaction is in equilibrium, ΔG = 0

QUESTION: 14

Identify the correct statement for change of Gibbs energy for a system (ΔG_{system}) at constant temperature and pressure: [2006]

Solution:

- 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.

QUESTION: 15

Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE? [2006]

Solution:

__We know that:__ ΔH = ΔE + PΔV

In the reactions, H_{2} + Br_{2 }→ 2HBr there is no change in volume or ΔV = 0.

So, ΔH = ΔE for this reaction.

QUESTION: 16

Consider the following reactions: [2007]

(i)

(ii)

(iii)

(iv)

Enthalpy of formation of H_{2}O (l) is

Solution:

**Reaction (ii) shows the formation of H _{2}O, and the X_{2} represents the enthalpy of formation of H_{2}O** 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.

QUESTION: 17

Given that bond energy of H–H and Cl–Cl is 430 kJ mol^{–1} and 240 kJ mol^{–1} respectively and ΔH_{f} for HCl is - 90 kJ mol^{–1}, bond enthalpy of HCl is: [2007]

Solution:

-90 = [1/2 x 430 + 1/2 x 240] - B.E. of HCl

∴ B.E. of HCl = 215 + 120 + 90 = 425 kJ mol^{–1}

QUESTION: 18

Which of the following are not state functions? [2008]

**(I)** q + w

**(II)** q

**(III)** w

**(IV)** H - TS

Solution:

► 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).

QUESTION: 19

For the gas phase reaction, which of the following conditions is correct? [2008]

PCl_{5}(g) ⇌ PCl_{3}(g) + Cl_{2}(g)

Solution:

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).

QUESTION: 20

Bond dissociation enthalpy of H_{2}, Cl_{2,} and HCl are 434, 242, and 431 kJ mol^{–1} respectively. Enthalpy of formation of HCl is: [2008]

Solution:

__The reaction of formation of HCl:__ H_{2 }+ Cl_{2} → 2HCI

⇒ Enthalpy of formation of 2HCl = - (862 - 676) = -186 kJ.

∴ Enthalpy of formation of HCl = -(186/2) kJ = -93 kJ

QUESTION: 21

The values of ΔH and ΔS for the reaction, C_{(graphite)} + CO_{2 (g) }→ 2CO_{(g)} are 170 kJ and 170 JK^{–1}, respectively. This reaction will be spontaneous at: [2009]

Solution:

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.

QUESTION: 22

From the following bond energies: [2009]

H – H bond energy: 431.37 kJ mol^{–1}

C = C bond energy: 606.10 kJ mol^{–1 }

C – C bond energy: 336.49 kJ mol^{–1 }

C – H bond energy: 410.50 kJ mol^{–1}

Enthalpy for the reaction, will be:

Solution:

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}

QUESTION: 23

Standard entropies of X_{2} , Y_{2} and XY_{3} are 60, 40 and 50 JK^{–1}mol^{–1 }respectively. For the reaction

to be at equilibrium, the temperature should be: [2010]

Solution:

ΔS for the reaction

ΔS = 50 – (30 + 60) = – 40 J

For equilibrium ΔG = 0 = ΔH – TΔS

QUESTION: 24

For vapor ization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63 kJmol^{–1} and 108.8 JK^{–1} mol^{–1}, respectively. The temperature when Gibbs energy change (ΔG) for this transformation will be zero, is: [2010]

Solution:

ΔH = 40630J mol ^{–1}

ΔS = 108.8JK^{–1} mol ^{–1}

∴ Correct choice : (d)

QUESTION: 25

Match List -I (Equations) with List-II (Type of processes) and select the correct option. [2010]

Options:

Solution:

∴ Correct choice : (d)

QUESTION: 26

The following two reactions are known : [2010]

The value of ΔH for the following reaction

Solution:

ΔH = –26.8 + 33.0 = + 6.2 kJ

∴ Correct choice : (a)

QUESTION: 27

If the enthalpy change for the transition of liquid water to steam is 30 kJ mol^{–1} at 27ºC, the entropy change for the process would be: [2011]

Solution:

Given ΔH = 30 kJ mol^{–1 }T = 273 + 27 = 300 K

= 100J mol^{–1} K^{–1}

QUESTION: 28

Enthalpy change for the reaction, 4H_{(g)} → 2H_{2 (g) } is - 869.6 kJ. The dissociation energy of H–H bond is: [2011]

Solution:

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

QUESTION: 29

Consider the following processes :

[2011M]

Solution:

To calculate ΔH operate 2 × eq. (1) + eq. (2) – eq. (3)

ΔH = 300 – 125 – 350 = – 175KJ/mol

QUESTION: 30

In which of the following reactions, standard entropy change (ΔS°) is positive and standard Gibb’s energy change (ΔG°) decreases sharply with increasing temperature? [2012]

Solution:

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.

QUESTION: 31

The enthalpy of fusion of water is 1.435 kCal/mol.The molar entropy change for the melting of ice at 0°C is: [2012]

Solution:

= 5.260 cal / mol-K

QUESTION: 32

Standard enthalpy of vapourisation Δ_{vap} H° for water at 100°C is 40.66 kJ mol^{–1}. The internal energy of vaporisation of water at 100°C (in kJ mol^{–1}) is : [2012]

(Assume water vapour to behave like an ideal gas).

Solution:

∆_{vap}H° = 40.66 kJ mol^{-1}

H_{2}O(l) ⇋ H_{2}O(g)

∆n_{g} = 1-0 = 1

∆H = ∆U + ∆n_{g}RT

∆U = ∆H-∆n_{g}RT

= 40,660 - 8.314×373

= 37558.878 J mol^{-1} or 37.56kJ mol^{-1}

QUESTION: 33

Equal volumes of two monoatomic gases, A and B, at the same temperature and pressure are mixed. The ratio of specific heats (C_{p}/C_{v}) of the mixture will be: [2012 M]

Solution:

QUESTION: 34

The Gibbs’ energy for the decomposition of Al_{2}O_{3} at 500°C is as follows :

The potential difference needed for the electrolytic reduction of aluminium oxide (Al_{2}O_{3}) at 500°C is at least : [2012 M]

Solution:

ΔG = -nFE°

for the reaction n

960 × 10^{3} = – 4 × 96500 × E°

E° = – 2.5 volt

So, it needed 2.5 volt for reduction

QUESTION: 35

When 5 liters of a gas mixture of methane and propane is perfectly combusted at 0°C and 1 atmosphere, 16 liters of oxygen at the same temperature and pressure is consumed. The amount of heat released from this combustion in kJ is: [NEET Kar. 2013]

ΔH_{comb} (CH_{4}) = 890 kJ mol^{–1}

ΔH_{comb} (C_{3}H_{8}) = 2220 kJ mol^{–1}

Solution:

2x + 5(5– x) = 16

⇒ x = 3L

∴ Heat released

QUESTION: 36

Three thermochemical equations are given below:

(i)

(ii)

(iii)

Based on the above equations, find out which of the relationship given below is correct? [NEET Kar. 2013]

Solution:

Applying Hess’s law, equation (i) can be obtained by adding equations (ii) and (iii).

∴ x = y + z

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