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 Multiple Choice QuestionsMultiple Choice Questions

101.

The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 250C; the heat of combustion (in kJ mol–1) of benzene at constant pressure will be (R = 8.314 JK–1 mol–1)

  • –3267.6

  • 4152.6

  • –452.46

  • 3260


A.

–3267.6

C6H6 (l) + 152 O2 (g)  6CO2 (g) +3H2O (l)  

Δng = 6 - 7.5
= -1.5 (change in gaseous mole)
ΔU or ΔE = - 3263.9 kJ
ΔH = ΔU + ΔngRT
Δng = - 1.5
R = 8.314 JK-1 mol-1
T = 298 K
So ΔH = -3263.9 + (-1.5) 8.314 x 10-3 x 298
= -3267.6 kJ
ΔH = Heat at constant pressure
ΔU/ΔE = Heat at constant volume
R = gas constant


102.

Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction?

  • A and D

  • A and B

  • B and C

  • C and D


B.

A and B

G° = - RT ln KH° - TS° = - RT ln K-H°RT+S°R = ln K

Therefore ln K vs 1/T the graph will be a straight line with slope equal to -H°R.Since reaction is
exothermic, therefore H° itself will be negative resulting in positive slope.


103.

Identify the invalid equation

  • H = ΣHproducts - ΣHreactants

  • H = U + pV

  • H°(reaction) = ΣH°(products bonds) - ΣH°(reactant bonds)

  • H = U + nRT


C.

H°(reaction) = ΣH°(products bonds) - ΣH°(reactant bonds)

Relation between heat of reaction (rH° ) and bond enthalpies of reactants and products is

rH° = ΣBEreactants - ΣBEproducts H°reaction = ΣH°product bonds - ΣH°reactant bonds


104.

For the process A (1, 0.05 atm, 32°C) → A (g, 0.05 atm, 32°C)

The correct set of thermodynamic parameters is

  • G = 0 and S = -ve

  • G = 0 and S = +ve

  • G = +ve and S = 0

  • G = -ve and S = 0


B.

G = 0 and S = +ve

(i) Since, n and T are constants. 

 G = 0

Hence, the system is equilibrium.

(ii) Since, P → decreases

 S → increases, i.e. it is +ve

Therefore, among all the options given, the correct answer is in option b, i.e. G = 0 and S = +ve.


105.

Two moles of an ideal monoatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.

  • (a) 195 K (b) 2.7 kJ

  • (a) 189 K (b) 2.7 kJ

  • (a) 195 K (b) –2.7 kJ

  • (a) 189 K (b) – 2.7 kJ


D.

(a) 189 K (b) – 2.7 kJ

In an adiabatic Process

TVγ-1 = ConstantOrT1V1γ-1 = T2V2γ-1 For monoatomic gasγ = 53(300)V2/3 = T2(2V)2/3 T2 = 300(2)2/3T2 = 189 K(final temperature)Change in internal energy,U = nf2RT= 232253(-111) = -2.7 kJ


106.

The first law of thermodynamics for isothermal process is

  • q = -W

  • U = W

  • U = qv

  • U = -qv


A.

q = -W

According to first law of thermodynamics,

U = q + W

where, U = internal energy; q = Heat and W = work done

For isothermal process,

 T = 0; U = 0

Therefore, q = -W