The amount of heat needed to raise the temperature of 60.0 g of aluminum from 35°C to 55°C would be -

(Molar heat capacity of Al is $24 \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}$)

$$\begin{gathered} 1. 1.07 \mathrm{J} \\ 2. 1.07 \mathrm{kJ} \\ 3. 106.7 \mathrm{kJ} \\ 4. 100.7 \mathrm{kJ} \end{gathered}$$

The enthalpy of formation of ${\mathrm{CO}}_{\left(\mathrm{g}\right)}, {\mathrm{CO}}_{2\left(\mathrm{g}\right)}, {\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)} , \mathrm{and} {\mathrm{N}}_2{\mathrm{O}}_{4\left(\mathrm{g}\right)}$are

–110 kJ ${\mathrm{mol}}^{-1}$, – 393 kJ ${\mathrm{mol}}^{-1}$, 81 kJ ${\mathrm{mol}}^{-\mathrm{1,}}$ and 9.7 kJ $mo{l}^{-1}$ respectively.

The value of ${∆}_{\mathrm{r}}\mathrm{H}$ for the reaction would be-

${\mathrm{N}}_2{\mathrm{O}}_{4\left(\mathrm{g}\right)}+3{\mathrm{CO}}_{\left(\mathrm{g}\right)}\to {\mathrm{N}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}+3{\mathrm{CO}}_{2\left(\mathrm{g}\right)}$

$$\begin{gathered} 1. -777.7 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 2. +777.7 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 3. +824.9 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 4. - 345.4 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

 ${\mathrm{N}}_2 + 3 {\mathrm{H}}_2 \to 2{\mathrm{NH}}_{3 ;} {∆}_{\mathrm{r}}{\mathrm{H}}^{°} = -92.4 \mathrm{kJ} {\mathrm{mol}}^{-1}$

The standard enthalpy of formation of ${\mathrm{NH}}_3$ gas in the above reaction would be-

$$\begin{gathered} 1. -92.4 \mathrm{J} {\mathrm{mol}}^{-1} \\ 2. -46.2 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 3. + 46.2 \mathrm{J} {\mathrm{mol}}^{-1} \\ 4. + 92.4 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

The standard enthalpy of formation of ${\mathrm{CH}}_3{\mathrm{OH}}_{\left(\mathrm{l}\right)}$ from the following data is -

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{OH}}_{\left(\mathrm{l}\right)}+ \frac{3}{2}{{\mathrm{O}}_2}_{\left(\mathrm{g}\right)} \to {{\mathrm{CO}}_2}_{\left(\mathrm{g}\right)} + 2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}; \\ {∆}_{\mathrm{r}}\mathrm{H}° = -726 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ {\mathrm{C(s)}}_{} + {{\mathrm{O}}_2}_{\left(\mathrm{g}\right)} \to {{\mathrm{CO}}_2}_{\left(\mathrm{g}\right)} ; \\ {∆}_{\mathrm{c}}\mathrm{H}° = -393 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ {{\mathrm{H}}_2}_{\left(\mathrm{g}\right)} + \frac{1}{2}{{\mathrm{O}}_2}_{\left(\mathrm{g}\right)} \to {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)} ; \\ {∆}_{\mathrm{f}}{\mathrm{H}}^{°} = -286 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

$$\begin{gathered} 1. -239 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 2. + 239 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 3. -47 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 4. +47 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

$$\begin{gathered} {∆}_{\mathrm{vap}}\mathrm{H}° \left({\mathrm{CCl}}_4\right) = 30.5 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ {∆}_{\mathrm{f}}\mathrm{H}° \left({\mathrm{CCl}}_4\right) = - 135.5 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ {∆}_{\mathrm{a}}\mathrm{H}° \left(\mathrm{C}\right) = 715.0 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ {∆}_{\mathrm{a}}\mathrm{H}° \left({\mathrm{Cl}}_2\right)\hspace{0.17em}= 242 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

The enthalpy change for the reaction

 ${\mathrm{CCl}}_4 \left(\mathrm{g}\right) \to \mathrm{C}\left(\mathrm{g}\right) + 4\mathrm{Cl} \left(\mathrm{g}\right)$ would be -

$$\begin{gathered} 1. 326 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 2. 1304 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 3. -328 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 4. -1304 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

For an isolated system with ∆U = 0, the ∆S value will be-

1. Positive

2. Negative

3. Zero

4. Not possible to define

For the reaction at 298 K,

2A + B → C

$∆\mathrm{H} = 400 \mathrm{kJ} {\mathrm{mol}}^{-1} \mathrm{and} ∆\mathrm{S} = 0.2 \mathrm{kJ} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$ .Reaction will become spontaneous at-

$$\begin{gathered} 1. 1500 \mathrm{K} \\ 2. 2000 \mathrm{K} \\ 3. 100 \mathrm{K} \\ 4. 1900\mathrm{K} \end{gathered}$$

For the reaction $2\mathrm{A} \left(\mathrm{g}\right) + \mathrm{B}\left(\mathrm{g}\right) \to 2\mathrm{D} \left(\mathrm{g}\right)$; $∆\mathrm{U}° = -10.5 \mathrm{kJ} \mathrm{and} ∆\mathrm{S}° = -44.1 \mathrm{J} {\mathrm{K}}^{-1}$ , the value of $∆\mathrm{G}°$ for the given reaction would be-

1. 1.6 J

2. - 0.16 KJ

3. 0.16 KJ

4. 1.6 KJ

The equilibrium constant for a reaction is 10. The value of $∆\mathrm{G}°$ will be-

($\mathrm{R} = 8.314 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1} ; \mathrm{T} = 300 \mathrm{K}$)

$$\begin{gathered} 1. -5.74 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 2. - 5.74 \mathrm{J} {\mathrm{mol}}^{-1} \\ 3. + 4.57 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 4. -57.4 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$

The reaction of cyanamide, ${\mathrm{NH}}_2\mathrm{CN} \left(\mathrm{s}\right)$ with dioxygen, was carried out in a bomb calorimeter, and ∆U was found to be $-742.7 \mathrm{kJ} {\mathrm{mol}}^{-1}$ at 298 K.

$$\begin{gathered} {\mathrm{NH}}_2\mathrm{CN} \left(\mathrm{s}\right) + \frac{3}{2}{\mathrm{O}}_2\left(\mathrm{g}\right) \to \\ {\mathrm{N}}_2 \left(\mathrm{g}\right) + {\mathrm{CO}}_2\left(\mathrm{g}\right) + {\mathrm{H}}_2\mathrm{O} \left(\mathrm{l}\right) \end{gathered}$$

The enthalpy change for the reaction at 298 K would be -

$$\begin{gathered} 1. -741.3 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 2. + 753.9 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 3. + 772. 7 \mathrm{kJ} {\mathrm{mol}}^{-1} \\ 4. -845. 1 \mathrm{kJ} {\mathrm{mol}}^{-1} \end{gathered}$$