Q: C ( ~s )+2 H _2( ~g ) CH _4( ~g ) ; H =-74.8 ~kJ ~mol ^ -1 Which of the following diagrams gives an accurate representation of the above reaction? [ R reactants; P products ]

For Exothermic reaction aligned & H=H_p-H_R=-v e \\ & H_R H_p aligned

Question 1

Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure. Processes 2 and 4 are adiabatic. w_1, w_2, w_3 and w_4 represent work done (in calories) in the processes 1, 2, 3

Accepted Answer

Using the IUPAC sign convention for thermodynamics, work done is given by w = - p dV. For the isothermal reversible process 1 (from volume V_1 to V_2 at constant temperature T_1): w_1 = -nRT_1 ( V_2 V_1 ) Given n = 1 mol and R = 2 cal K^ -1 mol^ -1 , we have: w_1 = -2T_1 ( V_2 V_1 ) For the isothermal reversible process 3 (from volume V_3 to V_4 at constant temperature T_2): w_3 = -nRT_2 ( V_4 V_3 ) = -2T_2 ( V_4 V_3 ) Adding the work done in these two processes: w_1 + w_3 = -2T_1 ( V_2 V_1 ) -

Question 2

Consider the following reaction : 2 A (g) + B (g) 2 D (g) U^ = -10 kJ mol^ -1 and S^ = -44 J K^ -1 at 298 K. Identify the correct option with G^ for the reaction and spontaneity of the reaction at 298

Accepted Answer

The change in the number of moles of gaseous species is: n_g = n_p - n_r = 2 - (2 + 1) = -1 The standard enthalpy change H^ is given by: H^ = U^ + n_g R T Substituting the given values: H^ = -10 + (-1) (8.31 10^ -3 ) 298 H^ = -10 - 2.47638 = -12.47638 kJ mol ^ -1 The standard Gibbs free energy change G^ is: G^ = H^ - T S^ G^ = -12.47638 - 298 (-44 10^ -3 ) G^ = -12.47638 + 13.112 = +0.63562 kJ mol ^ -1 Since G^ > 0, the reaction is non-spontaneous. Answer: +0.63568 kJ mol^ -1 , non-spontaneous

Question 3

Two moles of an ideal gas undergo free expansion from 10 L to 100 L at 300 K. The values of S_ system and S_ surroundings are (R is universal gas constant)

Accepted Answer

For the free expansion of an ideal gas, the external pressure is zero, so the work done W = 0. Since the process is isothermal at 300 K, the change in internal energy U = 0. From the first law of thermodynamics, q = U - W = 0. Since no heat is exchanged with the surroundings, the change in entropy of the surroundings is: S_ surroundings = q_ surr T = 0 Entropy is a state function, so the change in entropy of the system for an isothermal expansion is calculated using a reversible path between the

Question 4

At a certain temperature, T (K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then, change in internal energy of the system is :

Accepted Answer

According to the first law of thermodynamics: U = q + W Heat absorbed by the system, q = +500 J Work done by the system, W = -200 J Substituting the given values: U = 500 + (-200) = 300 J Answer: 300 J

Question 5

A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to N D . At 60 °C, the concentrations of both N and D are equal at equilibrium, and the stan

Accepted Answer

For the reversible thermal denaturation N D , the equilibrium constant is given by K = [ D ] [ N ] . At 60 ^ , the concentrations of N and D are equal, which means K = 1. The standard Gibbs free energy change is G^ = -RT K. Since K = 1, G^ = 0. Using the thermodynamic relation G^ = H^ - T S^ , we get: H^ = T S^ Given H^ = 666 kJ mol^ -1 and T = 60 + 273 = 333 K. S^ = H^ T = 666 333 = 2.0 kJ K^ -1 mol^ -1 Answer: 2.0

Question 6

C ( ~s )+2 H _2( ~g ) CH _4( ~g ) ; H =-74.8 ~kJ ~mol ^ -1 Which of the following diagrams gives an accurate representation of the above reaction? [ R reactants; P products ]

Accepted Answer

For Exothermic reaction aligned & H=H_p-H_R=-v e \ & H_R H_p aligned

Question 7

The standard heat of formation, in kcal / mol ^ of Ba ^ 2+ is : [Given : standard heat of formation of SO _4^ 2- ion (aq) =-216 kcal / mol , Standard heat of crystallisation of BaSO _4( ~s )=-4.5 kcal

Accepted Answer

aligned & H _ crys ( BaSO _4 )= H _ f ( BaSO _4 )- [ H _ f ( Ba ^ +2 )+ H _ f ( SO _4^ -2 ) ] \ & -4.5=[-349]- [ H _ f ( Ba ^ +2 )+(-216) ] \ & H _ f ( Ba ^ +2 )=-128.5 kCal ~mol ^ -1 aligned

Question 8

Choose the correct statement for the work done in the expansion and heat absorbed or released when 5 litres of an ideal gas at 10 atmospheric pressure isothermally expands into vacuum until volume is

Accepted Answer

Since it is isothermal, T =0 U = nC C _ v T =0 Since expansion is taking place against vacuum P_ ext =0 W=-P ex V=0 From first law of thermodynamics, U=q+W 0=q+0 q=0

Question 9

Match List-I with List-II: Choose the correct answer from the options given below:

Accepted Answer

Answer option index: 3

Question 10

For an endothermic reaction: (A) qp _ p is negative. (B) _r H is positive. (C) _r H is negative. (D) q _ p is positive. Choose the correct answer from the options given below:

Accepted Answer

For endothermic reactions, H _ r = positive (Heat is absorbed) H=E_ a(f) -E_ a(b)

Question 11

In which of the following processes entropy increases? A. A liquid evaporates to vapour. B. Temperature of a crystalline solid lowered from 130 K to 0 K . C. 2 NaHCO _ 3( ~s ) Na _2 CO _ 3( ~s ) + CO

Accepted Answer

When a liquid evaporates to vapour entropy increases 2 NaHCO _ 3( ~s ) Na _2 CO _ 3( ~s ) + CO _ 2( ~g ) + H _2 O _ ( g ) Number of gaseous product molecules increases so entropy increases. Cl _ 2( ~g ) 2 Cl _ ( g ) 1 mole Cl _ 2( ~g ) form 2 ~mol Cl _ ( g ) . So entropy increases.

Question 12

Match List I with List II. Choose the correct answer from the options given below:

Accepted Answer

(A) Isothermal process Temperature is constant throughout the process (B) Isochoric process Volume is constant throughout the process (C) Isobaric process Pressure is constant throughout the process (D) Adiabatic process No exchange of heat (q) between system and surrounding

Question 13

The work done during reversible isothermal expansion of one mole of hydrogen gas at 25^ C from pressure of 20 atmosphere to 10 atmosphere is (Given R =2.0 cal K ^ -1 ~mol ^ -1 )

Accepted Answer

aligned W_ rev, iso & =-2.303 n R T P_i P_f \ & =-2.303 1 2 298 2 \ & =-2.303 1 2 298 0.3 \ & =-413.14 calories aligned

Question 14

For the following reaction at 300 K A _2( ~g )+3 ~B _2( ~g ) 2 AB _3( ~g ) the enthalpy change is +15 kJ then the internal energy change is:

Accepted Answer

A _2( ~g )+3 ~B _2( ~g ) 2 AB _3( ~g ) n_ ( g ) = n _ ( P ) - n _ ( R ) =2-3-1=-2 H= U+ n_g R T 15 1000= U-2 8.314 300 U=15000+600 8.314 =15000+6 831.4 =15000+4988.4 U=19988.4 ~J

Thermodynamics (C) — Previous Year Questions & Solutions