AIEEE 2003 | Thermodynamics Question 172 | Chemistry

The internal energy change when a system goes from state A to B is 40 kJ/mole. If the system goes from A to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy?

> 40 kJ

< 40 kJ

Zero

40 kJ

AIEEE 2003

MCQ (Single Correct Answer)

-1

If at 298 K the bond energies of C - H, C - C, C = C and H - H bonds are respectively 414, 347, 615 and 435 kJ/mol, the value of enthalpy change for the reaction
H₂C = CH₂(g) + H₂(g) $$\to$$ H₃C - CH₃(g) at 298 K will be :

- 250 kJ

+ 125 kJ

- 125 kJ

+ 250 kJ

AIEEE 2003

MCQ (Single Correct Answer)

-1

In an irreversible process taking place at constant T and P and in which only pressure-volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria :

(dS)_{V, E} > 0, (dG)_{T, P} < 0

(dS)_{V, E} = 0, (dG)_{T, P} = 0

(dS)_{V, E} = 0, (dG)_{T, P} > 0

(dS)_{V, E} < 0, (dG)_{T, P} < 0

AIEEE 2003

MCQ (Single Correct Answer)

-1

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant K_c is :

- $$\Delta G$$ = RT ln K_c

$$\Delta G^o$$ = RT ln K_c

- $$\Delta G^o$$ = RT ln K_c

$$\Delta G$$ = RT ln K_c