Born-Haber cycle may be used to calculate from Chemistry Thermod

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531.

The maximum number of molecule is present in

  • 15 L of H2 gas at STP

  • 5 L of N2 gas at STP

  • 0.5 g of H2 gas

  • 10 g of O2 gas


532.

The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm, is

  • -6 J

  • -608 J

  • +304 J

  • -304 J


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533.

Born-Haber cycle may be used to calculate

  • electronegativity

  • mass number

  • oxidation number

  • electron affinity


D.

electron affinity

According to Born-Haber cycle the enthalpy of formation (Hf) of an ionic compound may be given as:

Hf = S+12D+I+E+U

where, I = ionisation energy; S = sublimation energy; E = electron affinity; D = dissociation energy and U = lattice energy of compound

Born-Haber cycle is used to determine the lattice energy of the compound. It also may be used to calculate electron affinity of an element.


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534.

The heat of formation for CO2 (g), H2O (l) and CH4 (g) are -400 kJ mol-1, -280 kJ mol-1 and -70 kJ mol-1 respectively. The heat of combustion of CH4 in kJ mol-1 is

  • 890

  • -160

  • -890

  • -90


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535.

C (s) + O2 (g) → CO2 (g); H = -94kcal

2CO (g) + O2 (g) → 2CO2 (g); H = -135 kcal.

The heat of formation of CO (g) is

  • -26.4 kcal

  • 41.2 kcal

  • 26.4 kcal

  • 229.2 kcal


536.

The entropy of a perfectly crystalline material is zero at 0°C. This is statement of

  • first law of thermodynamics

  • second law of thermodynamics

  • third law of thermodynamics

  • law of conservation of energy


537.

In any chemical reaction, a quantity that decrease to a minimum is

  • free energy

  • entropy

  • temperature

  • enthalpy


538.

Given, H2O (l) → H2O (g); H1 = +43 kJ

H2O (s) → H2O (l); H2 = +6.05 kJ. Calculate the enthalpy of sublimation of ice.

  • 49.75 kJ mol-1

  • 37.65 kJ mol-1

  • 43.7 kJ mol-1

  • 55.23 kJ mol-1


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539.

If standard enthalpies of formation of CaCl (s) (hypothetical) and that of CaCl(s) are -188 J mol-1 and -795 kJ mol-1 respectively, calculate the value of standard heat of reaction for the following  disproportionation reaction

2CaCl (s) → CaCl2 (s) + Ca (s)

  • -607 kJ mol-1

  • +607 kJ mol-1

  • -419 kJ mol-1

  • +419 kJ mol-1


540.

Heat formation of H2O is -188 kJ/ mol and H2O2 is 286 kJ/mol. The enthalpy change for the reaction; 2H2O2 → 2H2O + O2 is

  • 196 kJ

  • -196 kJ

  • 984 kJ

  • -984 kJ


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