Revisal Problems (Past 13 Years) JEE Main (Thermodynamics) - JEE MCQ

For the above reaction, 2NO(g) + O₂(g) → 2NO₂(g)

Select incorrect relation

At what temperature will the equilibrium constant for the formation of NOCI(g) be

K_eq = K_p = 1.0 x 10³

For the reaction at 298 K

Thus ΔG° for the given reaction is

Methane is a commercial source of H₂ through the reaction

Based on the following thermochemical equations (II to IV)

Q. ΔH of equation (I) is

Given,

Thus, for the reaction

Thus, for the reaction

Vapour pressure of a liquid is 10 mm at 300 K, 20 mm at 400 K. What is the vapour pressure at 500 K?

The combustion of hydrogen-oxygen mixture is used to produce very high temperatures (= 2500 °C) needed for certain types of welding operations.

Q. Quantity of heat, (in kJ) evolved when a 180 g mixture containing equal parts of H₂ and O₂ mass is burned, is

Due to dissolution of ammonium chloride in H₂O, temperature falls rapidly. Thus, 

The standard heat of combustion of benzoic acid, C₆H₅COOH(s), at 1 bar and 298 K is -329.3 kJ mol^-1. The standard heat of combustion at constant volume is

Bond dissociation energies of H—H, O=O, O—H and O—O are x₁, x₂, x₃, and x₄ respectively.

Thus, heat of formation of H₂O₂(g) is

Thus, melting point of NaCI (s) is

Quantity of work (in joules) done by the gas if it expands against a constant pressure of 0.980 atm and the change in volume (Δ/) is 25.0 L, is

The combustion of 1.0 g of sucrose (C₁₂H₂₂O₁₁) in a bomb calorimeter causes the temperature to rise from 25.0°C to 28.5°C. The heat of combustion of sucrose is (heat capacity of the calorimeter system is 4.90 kJ K^-1

A gas, while expanding absorbs 25 J of heat and does 243 J of work. Thus, ΔE (change in internal energy) for the gas is

1.0 L of liquid water freezes under a constant pressure of 1.0 atm and forms 1.1 L of ice.

Thus, work done is

In the following adiabatic expansion of 1 mole of CO₂ gas, temperature T is

For the dissociation of PCI₅(g)

Slope of the linear curve is such that θ = tan^-1 (-1.5)

Q. Thus, ΔH° is