Standard entropies of X₂, Y₂ and XY₃ are 60, 40 and 50 J K^-1mol^-1 respectively. For the reaction

$\frac{1}{2}{\mathrm{X}}_2 + \frac{3}{2}{\mathrm{Y}}_2 \stackrel{ }{\rightleftharpoons }{\mathrm{XY}}_3; ∆\mathrm{H} =-30 \mathrm{kJ},$
$\mathrm{to} \mathrm{be} \mathrm{at} \mathrm{equilibrium}, \mathrm{the} \mathrm{temperature} \mathrm{should} \mathrm{be}$
$1. 750\mathrm{K}$
$2. 1000\mathrm{K}$
$3. 1250\mathrm{K}$
$4. 500\mathrm{K}$

Given the following bond energies:

Based on the data given above, enthalpy change for the following reaction will be:
 

1. 1523.6 kJ mol^-1

2. -243.6 kJ mol^-1

3. -120.0 kJ mol^-1

4. 553.0 kJ mol^-1

The values of $∆$H and $∆$S for the given reaction are 170 kJ and 170 JK^–1, respectively.
\(\mathrm{C} \text { (graphite) }+\mathrm{CO}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}(\mathrm{g})\)
This reaction will be spontaneous at:

1. 710 K

2. 910 K

3. 1110 K

4. 510 K

On the basis of the following E$°$ values, the strongest oxidising agent is

[Fe(CN)₆]^4- $\to$[Fe(CN)₆]^3- + e^-         E$°$ = -0.35 V

 Fe²⁺ $\to$Fe³⁺ + e^-; E$°$ = -0.77 V

1. [Fe(CN)₆]^4-           

2. Fe²⁺

3. Fe³⁺                     

4. [Fe(CN)₆]^3-

Which of the following are not state functions?

(I) q + W                        (II) q

(III) W                           (IV) H-TS

1. (I) and (IV)                     

2. (II), (III) and (IV)

3. (I) , (II) and (III)               

4.  (II) and (III)

Consider the following reactions :

(i) ${\mathrm{H}}^{+}\left(\mathrm{aq}\right) + {\mathrm{OH}}^{-}\left(\mathrm{aq}\right) = {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) ∆\mathrm{H} = -{\mathrm{x}}_1 \mathrm{kJ} {\mathrm{mol}}^{-1}$

(ii) ${\mathrm{H}}_2\left(\mathrm{g}\right) + \frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right) = {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) ∆\mathrm{H} = -{\mathrm{x}}_2 \mathrm{kJ} {\mathrm{mol}}^{-1}$

(iii) ${\mathrm{CO}}_2\left(\mathrm{g}\right) + {\mathrm{H}}_2\left(\mathrm{g}\right) = \mathrm{CO}\left(\mathrm{g}\right) + {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) ∆\mathrm{H} = -{\mathrm{x}}_{3 }\mathrm{kJ} {\mathrm{mol}}^{-1}$

(iv) ${\mathrm{C}}_2{\mathrm{H}}_5\left(\mathrm{g}\right) + \frac{5}{2}{\mathrm{O}}_2\left(\mathrm{g}\right) = 2{\mathrm{CO}}_2\left(\mathrm{g}\right) + {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) ∆\mathrm{H} = -{\mathrm{x}}_4 \mathrm{kJ} {\mathrm{mol}}^{-1}$

Enthalpy of formation of H₂O(l) is:

1. -x₂ kJ mol^-1 

2. +x₃ kJ mol^-1

3. -x₄ kJ mol^-1

4. -x₁ kJ mol^-1

Given those bond energies of H-H and Cl-Cl are 430 kJ mol^-1 and 240 kJ mol^-1 respectively and ΔH_f for HCI is -90 kJ mol^-1. Bond enthalpy of HCl is:

1. 290 kJ mol^-1

2. 380 kJ mol^-1

3. 425 kJ mol^-1

4. 245 kJ mol^-1

Identify the correct statement for change of Gibbs energy for a system ($∆$G_system) at constant temperature and pressure:

(1) If $∆$G_system > 0, the process is spontaneous

(2) If $∆$G_system = 0, the system has attained equilibrium

(3) If $∆$G_system = 0, the system is still moving in a particular direction

(4) If $∆$G_system < 0, the process is not spontaneous

The enthalpy and entropy change for the reaction :

Br₂ (l) + Cl₂ (g)$\to$ 2BrCl (g)

are 30 kJ mol^-1 and 105 J K^-1 mol^-1 respectively.

The temperature at which the reaction will be in equilibrium is :

Consider the reactionat 300K

H₂₍₉₎ + Cl₂₍₉₎ →2HCI_(g), ΔH° = — 185 KJ

If 3 mole of H₂ completely react with 3 mol of Cl₂ to form Cl, $∆$U° of the reaction will be

(1) Zero

(2) –185 KJ

(3) -555 KJ

(4) None