What is the entropy change (in JK^–1 mol^–1) when one mole of ice is converted into water at 0 ºC? (The enthalpy change for the conversion of ice to liquid water is 6.0 KJ mol^–1 at 0 ºC)

1.	20.13	2.	2.013
3.	2.198	4.	21.98

The formation of a solution from two components can be considered as:

The solution so formed will be ideal if:
1. ∆H_Soln = ∆H₁ + ∆H₂ + ∆H₃

2. ∆H_Soln = ∆H₁ + ∆H₂ – ∆H₃

3. ∆H_Soln = ∆H₁ – ∆H₂ – ∆H₃

4. ∆H_Soln = ∆H₃ – ∆H₁ – ∆H₂

The molar heat capacity of water at constant pressure, C, is 75 JK^–1 mol^–1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of the water is:

$\frac{1}{2}{N_2}_{ \left(\mathrm{g}\right)} + \frac{1}{2}{O}_{2 \left(\mathrm{g}\right)} \to N{O}_{\left(\mathrm{g}\right)} ;$
${∆}_{\mathrm{r}}H° = 90 kJ mo{l}^{-1}$
$N{O}_{\mathit{\left(}g\mathit{\right)}} + \frac{1}{2}{O_2}_{\mathit{\left(}g\mathit{\right)}} \to N{O_{\mathit{2}}}_{\mathit{\left(}g\mathit{\right)}}\hspace{0.17em};$
${∆}_{\mathrm{r}}\mathrm{H}° = -74 \mathrm{kJ} {\mathrm{mol}}^{-1}$

The thermodynamic stability of NO_(g) based on the above data is:

1. Less than NO_2(g) 

2. More than NO_2(g)

3. Equal to NO_2(g)

4. Insufficient data

The entropy change in the surroundings when 1.00 mol of H₂O_(l) is formed under standard conditions is:

∆_fH^θ = –286 kJ mol^–1 

1. 952.5 J mol^-1

2. $979.7$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$

3. $949.7$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$

4. $959.7$ $\mathrm{J}$ ${\mathrm{mol}}^{-1}$

For the graph given below, it can be concluded that work done during the process shown will be-

Consider the following graph.

The work done, as per the graph above, is:

Consider the following diagram for a reaction $\mathrm{A}\to \mathrm{C}$. 

The nature of the reaction is-

1. Exothermic

2. Endothermic

3. Reaction at equilibrium

4. None of the above

What is the nature of the reaction depicted in the given diagram for A→C?

The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm^–3, respectively. If the standard free energy difference (∆Gº) is equal to 1895 J mol^–1, the pressure at which graphite will be transformed into diamond at 298 K is:

1. 11.08$\times {10}^8$ $\mathrm{Pa}$

2. $9.92\times {10}^7$ $\mathrm{Pa}$

3. $9.92\times {10}^6$ $\mathrm{Pa}$

4. 11.08$\times {10}^5$ $\mathrm{Pa}$