$2Zn+O_2\to 2ZnO$ $$ $$ $$ $∆G°=-616$ $J$
$2Zn+S_2\to 2ZnS$ $$ $$ $$ $$ $∆G°=-293$ $J$
$S_2+2O_2\to 2S{O}_2$ $$ $$ $$ $$ $∆G°=-408J$

$∆G°$ for the following reaction is:

$2ZnS+3O_2\to 2ZnO+2S{O}_2$

1.	-731 J	2.	-1317 J
3.	-501 J	4.	+731 J

At 27ºC latent heat of fusion of a compound is 2930 J/mol. Entropy change is:

1. 9.77 J/mol K

2. 10.77 J/mol K

3. 9.07 J/mol K

4. 0.977 J/mol K

For the reaction
C₂H₅OH(l) + 3O₂(g) →2CO₂(g) + 3H₂O(l) which one is true:

In a closed insulated container a liquid is stirred with a paddle to increase the temperature. The correct option regarding this among the following is:

1. ∆E = W ≠ 0, q = 0

2. ∆E = W = q ≠ 0

3. ∆E = 0, W = q ≠ 0

4. W = 0 , ∆E = q ≠ 0

2 mole of an ideal gas at 27ºC temp. is expanded reversibly from 2 lit. to 20 lit. Find entropy change (R = 2 cal/mol K): 

1. 92.1

2. 0

3. 4

4. 9.2

Heat of combustion ∆Hº for C(s), H₂(g) and CH₄(g) are – 94, – 68 and – 213 Kcal/mol. ∆Hº for C(s) + 2H₂(g) → CH₄ (g) is:

When 1 mol gas is heated at constant volume, the temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. The correct statement among the following is:

1.  q = w = 500 J, ∆U = 0
2.  q = ∆U = 500 J, w = 0
3.  q =0, w = 500 J, ∆U = 0
4.  ∆U = 0, q = w = – 500 J

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$\times {10}^5$ $\mathrm{Pa}$

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)

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