Thermodynamics

The first law of thermodynamics for isothermal process is

• q = -W

• $∆$U = W

• $∆$U = q_v

• $∆$U = -q_v

Identify the invalid equation

• $∆\mathrm{H} = {\mathrm{\Sigma H}}_{\mathrm{products}} - {\mathrm{\Sigma H}}_{\mathrm{reactants}}$

• $∆\mathrm{H} = ∆\mathrm{U} + \mathrm{p}∆\mathrm{V}$

• $∆\mathrm{H}{°}_{\left(\mathrm{reaction}\right)} = \mathrm{\Sigma H}{°}_{(\mathrm{products} \mathrm{bonds})} - \mathrm{\Sigma H}{°}_{(\mathrm{reactant} \mathrm{bonds})}$

• $∆\mathrm{H} = ∆\mathrm{U} + ∆\mathrm{nRT}$

For the process A (1, 0.05 atm, 32°C) → A (g, 0.05 atm, 32°C)

The correct set of thermodynamic parameters is

• $∆$G = 0 and $∆$S = -ve

• $∆$G = 0 and $∆$S = +ve

• $∆$G = +ve and $∆$S = 0

• $∆$G = -ve and $∆$S = 0

Thermal decomposition of ammonium dichromate gives

• N₂, H₂O and Cr₂O₃

• N₂, NH₃ and CrO

• (NH₄)₂CrO₄ and H₂O

• N₂, H₂O and CrO₃

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

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

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

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

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

Mathematical equation of first law of thermodynamics for isochoric process is

• ΔU = q_v

• -ΔU = q_v

• q= -W

• ΔU = W