The ratio of heats liberated at 298 K from the combustion of one

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 Multiple Choice QuestionsMultiple Choice Questions

511.

Based on the first law of thermodynamics, which one of the following is correct?

  • For an isochoric process = ΔE = - q

  • For an adiabatic process = ΔE =-w

  • For an isothermal process = q = + w

  • For a cyclic process q = - w


512.

For the reversible reaction, A (s) + B (g)  C (g) + D (g), G° = -350 kJ, which one of the following statements is true?

  • The reaction is thermodynamically non-feasible

  • The entropy change is negative

  • Equilibrium constant is greater than one

  • The reaction should be instantaneous


513.

The value of entropy of solar system is

  • Increasing

  • decreasing

  • constant

  • zero


514.

Which of the following statements is true?

  • The total entropy of the universe is continuously decreasing

  • The total energy of the universe is continuously decreasing

  • The total energy of the universe remains constant

  • The total entropy of the universe remains constant


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515.

A gas expands from a volume of 1m3 to a volume of 2m3 against an external pressure of 105Nm-2. The work done by the gas will be

  • 102 kJ

  • 102 J

  • 103 J

  • 105 kJ


516.

Given thermochemical equation, 2H2 (g) + O2(g) → 2H2O,ΔH = -571.6 kJ. Heat of decomposition of water is

  • -571.6 kJ

  • +571.6 kJ

  • -1143.2 kJ

  • + 285.8 kJ


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517.

The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas obtained from kg of coke is (Assume coke to be 100% carbon). (Given : enthalpies of combustion of CO2, CO and H2 as 393.5 kJ, 285 kJ, 285 kJ respectively all at 298 K)

  • 0.79 : 1

  • 0.69 : 1

  • 0.86 : 1

  • 0.96 : 1


B.

0.69 : 1

Number of moles in 1 kg coke = 100012 = 83.33 mol

(i) For the combustion of 1 kg of coke

C + O2 → CO2H = 393.5 kJ

Heat liberated from 1 mole coke = 393.5 kJ

 Heat liberated from 83.33 mole coke = (393.5 × 83.33) kJ

(ii) For the burning of water gas

C + H2O → CO + H2

CO + H+ O2 → CO2 + H2O

H = 285 + 285 = 570 kJ

 Ratio of heat liberated from burning of water gas obtained from 1 kg of coke

H2 = (570 × 83.33) kJ

Therefore, required ratio = H1.H2

                                    = 393.5 × 570

                                   = 0.69 : 1


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518.

The process is spontaneous at the given temperature, if

  • ΔH is +ve and ΔS is -ve

  • ΔH is -ve and ΔS is +ve

  • ΔH is +ve and ΔS is +ve

  • ΔH is +ve and ΔS is equal to zero.


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519.

For an ideal binary liquid mixture.

  • Smix = 0; Gmix = 0

  • Hmix = 0; Smix < 0

  • Vmix = 0; Gmix > 0

  • Smix > 0; Gmix < 0


520.

The correct statement regarding entropy is

  • at absolute zero temperature, entropy of a perfectly crystalline solid is zero

  • at absolute zero temperature, the entropy of a perfectly crystalline substance is positive

  • at absolute zero temperature, the entropy of all crystalline substances is zero

  • at 0°C, the entropy of a perfect crystalline solid is zero


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