An ideal gas expands in volume from \(1×10^{–3} m^3\) to \(1×10^{–2} m^3\) at 300 K against a constant pressure of 1×10⁵ Nm^-2. The work done is:

1.	–900 J	2.	–900 kJ
3.	270 kJ	4.	900 kJ

The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJ mol^-1 respectively. The enthalpy of the formation of carbon monoxide per mole is:

1. 110.5 kJ

2. 676.5 kJ

3. -676.5 kJ

4. -110.5 kJ

The enthalpy changes for the following processes are listed below : 

Cl₂ (g) → 2Cl(g), 242.3 kJ mol^-1 

I₂ (g) → 2I (g), 151.0 kJ mol^-1 

ICI (g) → I(g) + Cl(g), 211.3 kJ mol^-1 

I₂ (s) → I₂ (g), 62.76 kJ mol^-1

Given that the standard states for iodine and chlorine  are I₂ (s) and Cl₂ (g), the standard enthalpy of formation of ICI (g) is : 

1. -14.6 kJ mol^-1 

2. -20.8 kJ mol^-1 

3. +16.8  kJ mol^-1 

4. +244.8  kJ mol^-1

\((\Delta H-\Delta U)\)  for the formation of carbon monoxide (CO) from its elements at 298 K is : 

(R = 8.314 JK^-1 mol^-1)

1. -1238.78 J mol^-1 

2. 1238.78 J mol^-1 

3. -2477.57 J mol^-1 

4. 2477.57 J mol^-1

If a reaction is non-spontaneous at the freezing point of water but is spontaneous at the boiling point of water, then:

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^–1, the value of enthalpy change for the reaction at 298 K will be:

\(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3}(g)\)

1. +250 kJ 

2. –250 kJ 

3. +125 kJ 

4. –125 kJ 

The internal energy change when a system goes from state A to B is 40 k J/mol . 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? 

1. 40 kJ 

2. > 40kJ 

3. < 40 kJ 

4. zero

Assuming that water vapor is an ideal gas, the internal energy change (∆U) when 1 mol of water is vaporized at 1 bar pressure and 100°C, will be:

(Given: Molar enthalpy of vaporization of water at 1 bar and 373 K = 41 kJ mol^–1 and R = 8.3 J mol^–1 K^–1) 

1. 4.100 kJ mol^–1

2. 3.7904 kJ mol^–1

3. 37.904 kJ mol^–1

4. 41.00 kJ mol^–1

Standard entropy of ${\mathrm{X}}_2,{\mathrm{Y}}_2$ and ${\mathrm{XY}}_3$ are $60,40$ and $50\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1},$ respectively. For the reaction, $\frac{1}{2}<mspace\; width="1em"></mspace>{\mathrm{X}}_2+\frac{3}{2}<mspace\; width="1em"></mspace>{\mathrm{Y}}_2<mspace\; width="1em"></mspace>\to <mspace\; width="1em"></mspace>{\mathrm{XY}}_3∆\mathrm{H}=-30<mspace\; width="1em"></mspace>\mathrm{kJ},$ to be at equilibrium, the temperature will be

1. 500 K

2. 750 K

3. 1000 K

4. 1250 K