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Thermodynamics

JEE Main 2024 (Online) 30th January Evening ShiftNumericalEasy📈 Topic in 10% papers+4
Ques. 1JEE Main 2024, 30 January (Evening Shift)

Two reactions are given below: 2Fe(s) + 32O2(g) Fe2O3(s), H = -822 kJ/mol C(s) + 12O2(g) CO(g), H = -110 kJ/mol Then enthalpy change for following reaction 3C(s) + Fe2O3(s) 2Fe(s) + 3CO(g) is kJ/mol.

Answer

492

Solution

Hess's law: required reaction = 3 × (ii) − (i)

H = 3(-110) - (-822)

= -330 + 822

= +492 kJ/mol

Answer: 492

Keep practising — Thermodynamics

Ques. 2JEE Main 2024, 29 January (Morning Shift)

Which of the following is not correct?

  1. G is positive for a spontaneous reaction
  2. G is positive for a non-spontaneous reaction
  3. G is zero for a reversible reaction
  4. G is negative for a spontaneous reaction
Ques. 3JEE Main 2024, 29 January (Evening Shift)

Standard enthalpy of vapourisation for CCl4 is 30.5 kJ mol-1. Heat required for vapourisation of 284 g of CCl4 at constant temperature is kJ. (Given molar mass in g mol-1; C = 12, Cl = 35.5)

Ques. 4JEE Main 2024, 27 January (Morning Shift)

If three moles of an ideal gas at 300 K expand isothermally from 30 dm3 to 45 dm3 against a constant opposing pressure of 80 kPa, then the amount of heat transferred is J.

Ques. 5JEE Main 2024, 27 January (Evening Shift)

For a certain thermochemical reaction M N at T = 400 K, = 77.2 kJ mol-1, = 122 J K-1, log equilibrium constant (log K) is − 10-1.

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