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**Q.1. For the reactionA(l) → 2B(g)**

Hence ΔG in kcal is ______. (2020)

Ans.

For the reaction

A(l) → 2B(g)

Δn

From the relation

ΔH = ΔU + Δn

Substituting the values in Eq. (1), we get

ΔH = 2.1 × 10

We know

ΔG = ΔH − T ΔS ...(2)

Substituting the values in Eq. (2), we get

ΔG = 3300 - 300 × 20

= -2700 cal = - 2.7 kcal

Ans.

We know,

Standard heat of formation of ethane is Δ

Given:

...(1)

...(2)

...(3)

On applying algebraic operations on equation, that is, 2 × Eq. (3) + 3 × Eq. (2) – Eq. (1), We get,

Ans.

The magnitude of word done undergo a reversible expansion along the path ABC is the are of the trapezium ABC.

Thus, the word done is

Ans.

We know

ΔU = nC

Substituting the given values in Eq. (1), we get

(1) is x = y

(2) Does not exist

(3) is x > y

(4) is x < y

Ans.

Given, enthalpy of atomization of B

From the above two reactions

x = Δ

⇒ x > y.

(1) Both ΔS and S are functions of temperature.

(2) Both S and ΔS are not functions of temperature.

(3) S is not a function of temperature but ΔS is a function of temperature.

(4) S is a function of temperature but ΔS is not a function of temperature.

Ans.

Both entropy (S) and change in entropy (ΔS) is the function of temperature.

(1)

(2)

(3)

(4) |

Ans.

⇒

|w| = nRT (InV

|w| = nRT In V

y = mx + c

So, slope of curve 2 is more than curve 1 and intercept of curve 2 is more negative than curve 1.

(log 273 = 2.436, log 373 = 2.572, log 383 = 2.583) (2019)

(1) 7.90 kJ kg

(2) 2.64 kJ kg

(3) 8.49 kJ kg

(4) 9.26 kJ kg

Ans.

As we know,

Also,

Also,

Now,

∴ Total entropy change

ΔS = 9.26 kJ kg

(1) 20 K

(2) 12 K

(3) 5 K

(4) 4 K

Ans.

ΔH = 200 J mol

ΔS = 40 JK

For spontaneous reaction,

ΔG < 0

ΔH - TΔS < 0; ΔH < TΔS

5 < T

So, minimum temperature is 5 K.

(1) Dissociation of CaSO

(2) Sublimation of dry ice

(3) Dissolution of iodine in water

(4) Synthesis of ammonia from N

Ans.

In the process of synthesis of ammonia from N

N

(1) 3/2 K

(2) 2 K

(3) 2/3 K

(4) 1 K

Ans.

We know that,

w = -P

w = -4 Nm

w = 16 Nm ⇒ 16 J

For isothermal compression,

ΔU = q + w

⇒ q = -w = -16 J (∵ ΔU = 0 for isothermal process)

From calorimetry,

Heat given = n C ΔT

∴ Change in temperature, ΔT = 2/3 K

(1)

(2)

(3)

(4)

**Ans.** (1)

Final temperature**Q.13. For the chemical reaction X ⇌ Y, the standard reaction Gibbs energy depends on temperature T (in K) asThe major component of the reaction mixture at T is: (2019)(1) Y if T= 300 K(2) Y if T= 280 K(3) X if T = 350 K(4) X if T = 315 KAns. **(4)

At 315 K; ΔGº

ΔG° = 120 - 118.125 = positive

Since ΔG° is positive then K

(1) Exothermic if B < 0

(2) Endothermic if A > 0

(3) Endothermic if A < 0 and B > 0

(4) Exothermic if A > 0 and B < 0

Ans.

MgO (s) + C(s) → Mg (s) + CO (g)

For a reaction to be spontaneous

ΔG < 0

ΔH° - TΔS° < 0

T > 2480.3 K

(1)

(2)

(3)

(4)

Ans.

ΔG° = A - BT

A and B are non-zero constants

∴ ΔG° = ΔH° - TΔS° = A - BT

∴ Reaction will be endothermic if ΔG° > 0.

Hence, A > O and B < O.

(1) x = y + z

(2) z = x + y

(3) y = 2z - x

(4) x = y - z

Ans.

(ΔT

K

m

(1) 62 kJ

(2) 16 kJ

(3) 21 kJ

(4) 13 kJ

Ans.

T

C

The relation between ΔH and C is

After putting all variable values in eq. (i)

= 3[16100 + 4550] = 3 × 20650 = 61950 J

= 61.95 kJ

= 62 kJ

(1) Cyclic process : q = -w

(2) Adiabatic process: ΔU= -w

(3) Isochoric process: ΔU = q

(4) Isothermal process: q = -w

Ans.

From first law of thermodynamics, ΔU = q + w

For adiabatic process, q = 0

∴ ΔU = w

For isothermal process, ΔU = 0

For cyclic process, ΔU = 0

For isochoric process, w = 0

Ans.

ΔU = nCv ΔT = 5 × 28 × 100 = 14 KJ

Δ(PV) = nR (T

(a) q + w

(b) q

(c) w

(d) H-TS

(1) (B) and (C)

(2) (B), (C) and (D)

(3)(A) and (D)

(4) (A), (B) and (C)

Ans.

We know that heat and work are not state functions but q + w = ΔU is a state function. H -TS (i.e. G) is also a state function.

Identify the incorrect statement. (2019)

(1) Activation enthalpy to form C is 5kJ mol

(2) C is the thermodynamically stable product.

(3) D is kinetically stable product.

(4) Formation of A and B from C has highest enthalpy of activation.

Ans.

As we can see from the graph that activation enthalpy to form D from A+B is 15 - 5 = 10 kJ mol

(1) -12

(2) -8

(3) 8

(4) 12

Ans.

w= 10 kJ

q = -2 kJ

ΔU = q + w = -2 + 10 = 8 kJ

(1) ΔH < 0 and ΔS < 0

(2) ΔH > 0 and ΔS < 0

(3) ΔH < 0 and ΔS > 0

(4) ΔH > 0 and ΔS > 0

Ans.

A reaction is spontaneous if ΔG

ΔG

A reaction will be spontaneous at all temperatures if ΔH

(1) -4 RT

(2) -3RT

(3) 4RT

(4) 3 RT

Ans.

ΔH - ΔU = Δn

Δn

∵ Δn

∴ ΔH - ΔU = -4RT

(1) -9.0

(2) +10.0

(3) -0.9

(4) -2.0

Ans.

w = - PΔV

= -(1 bar) x (9 L)

= -(10

= - 9 x 10

= - 900 J

= - 0.9 kJ

(1) 2.85

(2) 5.7

(3) 22.8

(4) 11.4

Ans.

I

Heat of reaction depend upon temperature i.e., it varies with temperature, as given by Kirchoff's equation,

where ΔC

∴ ΔC

ΔH

ΔH

(1) ΔG° < 0, K > 1

(2) ΔG° = 0, K = 1

(3) ΔG° > 0, K < 1

(4) ΔG° < 0, K < 1

Ans.

ΔGº = -RT Ink

∴ If K > 1 then ΔGº < 0

If K < 1 then ΔGº > 0

If K = 1 then ΔGº = 0

(1) A and B

(2) B and C

(3) C and D

(4) A and D

Ans.

ΔGº = -RT In K

ΔHº - TΔSº = -RT In K

Therefore In K vs 1/T graph will be a straight line with slope equal to Since reaction is exothermic, therefore ΔH

(2) -452.46

(3) 3260

(4) -3267.6

Ans.

Δng = 6-7.5 = -1.5

ΔH = ΔU + ΔngRT

= -3263.9 - 1.5 x 8.134 x 10

= -3267.6

ΔU

W

Heat absorbed by the system during process CA is:

(2) -5 kJ mol

(3) +5 KJ mol

AB → isobaric

BC → Isochoric

CA → not defined

ΔU

= 2 − 5 = −3

ΔU

=− 3 − 5= − 8 kJ

ΔU

= Q + W

8 = Q + 3

Q = +5 kJ

(1) C(diamond) → C(graphite)

(2) N

(3) N

(4) H

Ans.

N

for isothermal process

Therefore, ΔS < 0

(1) Isochoric work

(2) Isobaric work

(3) Adiabatic work

(4) Isothermal work

Ans.

For adiabatic process, q = 0

∴ As per 1

ΔU = W

The correct statement for the reaction is: (2017)

(1) ΔH = ΔU = 0

(2) |ΔH| < |ΔU|

(3) |ΔH| > |ΔU|

(4) ΔH = ΔU ≠ 0

Ans.

For the reaction A(g) → A(1)

The change in the number of moles of gaseous species Δn = 0 - 1 = -1

The enthalpy change ΔH = -3RT

ΔH = ΔU + ΔnRT

-3RT = ΔU + (-1)RT

The change in internal energy ΔU = -2RT

Hence, |ΔH| > |ΔU|

(1) 6.00 kJ mol

(2) 5.81 kJ mol

(3) 5.44 kJ mol

(4) 6.56 kJ mol

Ans.

In order to calculate the enthalpy change for H

(a) Energy change of 1 mol, H

(b) Energy change of 1 mol, H

(c) Energy change of 1 mol, Ice(s), at 0ºC → 1 mol, Ice(s), -5

Total ΔH = C

= (75.3 J mol

ΔH = -6.56 kJ mol

So, ΔH = 6.56 kJ

(1) 6J of the work will be done by the gas

(2) 6J of the work will be done by the surrounding on gas.

(3) 10 J of the work will be done by the surrounding on gas.

(4) 10 J of the work will be done by the gas

Ans.

The heat of formation (in kJ) of carbon monoxide per mole is: (2016)

(1) 110.5

(2) 676.5

(3) -676.5

(4) -110.5

Ans.

(1) Both ΔH and ΔS are negative.

(2) Both ΔH and ΔS are positive.

(3) ΔH is positive while ΔS is negative.

(4) ΔH is negative while ΔS is positive.

Ans.

ΔG=ΔH−T. ΔS

If ΔH & ΔS are both positive, then ΔG may be negative at height temperature. Hence reaction becomes spontaneous at higher temperature.

2H

(1) 498.00

(2) 62.25

(3) 124.50

(4) 249.00

Ans.

2H

W = –P

∵ 100 mol H

⇒ W = –(50)(8.3)(300)J

= - 124500 J

W = – 124.5 kJ

⇒ Work done by O

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