The standard enthalpies of combustion of C6H6 (l), C(graphite) an

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 Multiple Choice QuestionsMultiple Choice Questions

91.

Thermal decomposition of ammonium dichromate gives

  • N2, H2O and Cr2O3

  • N2, NH3 and CrO

  • (NH4)2CrO4 and H2O

  • N2, H2O and CrO3


92.

The enthalpy change for a reaction at equilibrium is -20.5 kJ mol-1. Then the entropy change for this equilibrium at 410 K is

  • +50 JK-1mol-1

  • +55 JK-1mol-1

  • +75 JK-1mol-1

  • -50 JK-1mol-1


93.

The enthalpy of combustion of glucose (mol.wt.: 180 g mol-1) is -2840 kJ mol-1. Then the amount of heat evolved when 0.9 g ofglucose is burnt, will be

  • 14.2 kJ

  • 142 kJ

  • 28.4 kJ

  • 1420 kJ


94.

The entropy of vaporisation of a liquid is 58 JK-1 mol-1. If 100 g of its vapour condenses at its boiling point of 123° C, the value of entropy change for the process is

  • -100 JK-1

  • 100 JK-1

  • -123 JK-1

  • 123 JK-1


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95.

At constant external pressure of one atmosphere, 4 moles of a metallic oxide MO2 undergoes complete decomposition at 227°C in an open vessel according to the equation

2MO2 (s) → 2MO (s) + O2 (g)

The work done by the system in kJ is (R = 8.3 kJ mol-1)

  • -16.6

  • -24.9

  • -33.2

  • -4.15


96.

A certain reaction has a H of 12 kJ and a S of 40 JK-1. The temperature above which the reaction becomes spontaneous is

  • 27°C

  • 27 K

  • 300°C

  • 30 K


97.

Mathematical equation of first law of thermodynamics for isochoric process is

  • ΔU = qv

  • -ΔU = qv

  • q= -W

  • ΔU = W


98.

The criterion for a spontaneous process is

  • ΔG >0 

  • ΔG < 0

  • ΔG = 0

  • ΔStotal < 0


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99.

Choose the reaction in which H is not equal to U

  • C(graphite) + O2(g) → CO2(g)

  • C2H4(g) + H2(g) → C2H6(g)

  • 2C(graphite) + H2(g) → C2H2(g)

  • H2(g) + I2(g) → 2HI(g)


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100.

The standard enthalpies of combustion of C6H6 (l), C(graphite) and H2 (g) are respectively -3270 kJ mol-1, -394 kJ mol-1 and -286 kJ mol-1. What is the standard enthalpy of formation of C6H6 (l) in kJ mol-1?

  • -48

  • +48

  • -480

  • +480


B.

+48

Given, 

(i) C6H6152O2 → 6CO2 + 3H2O; H = -3270 kJ mol-1

(ii) C(graphite) + O2 → CO2H = -394 kJ mol-1

(iii) H212O2 → H2O; H = -286 kJ mol-1

On multiplying equation (ii) by 6 and equation (iii) by 3, we get,

(iv) 6C (s) + 6O2 → 6CO2H = -394 × 6 kJ mol-1 = -2364 kJ mol-1

(v) 3H232O2 → 3H2O; H = -286 × 3 kJ mol-1 = -858 kJ mol-1 

On inverting equation (i), we get,

(vi) 6CO2 + 3H2O → C6H6152O2H = +3270 kJ mol-1

On adding equation (iv), (v) and (vi),

6C (s) + 3H2 → C6H6H = +3270 + (-2364 - 858) = + 48 kJ mol-1

Thus, the standard enthalpy of formation of C6H6 is +48 kJ mol-1.


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