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

1.	∆H = ∆E – RT	2.	∆H = ∆E + RT
3.	∆H = ∆E + 2RT	4.	∆H = ∆E – 2RT

Unit of entropy is:

1. JK^–1 mol^–1

2. J mol^–1

3. J^–1K^–1 mol^–1

4. JK mol^–1

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:

For the given reaction 

$2H_2{O}_2\left(l\right)$ $\to$ $2H_{2}O\left(l\right)$ $+$ $O_2\left(g\right)$, the heat of formations of ${\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)$ $$ $\mathrm{and}$ ${\mathrm{H}}_2\mathrm{O}$ $\left(\mathrm{l}\right)$ $$are -188 kJ/mol & -286 KJ/mol respectively. The change in the enthalpy of the reaction will be:

1.  – 196 kJ/mol

2.  + 196 kJ/mol

3.  + 948 kJ/mol

4.  – 948 kJ/mol

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

Enthalpy  of $C{H}_4$ $+$ $\frac{1}{2}{O}_2$ $\to$ $C{H}_{3}OH$ is negative. If the enthalpy of combustion of $C{H}_4$ $and$ $C{H}_{3}OH$ are x and y respectively, then which relation is correct:

For the reaction:

\(C_{3} H_{8} \left(\right. g \left.\right) + 5 O_{2} \left(\right. g \left.\right) \rightarrow 3 CO_{2} \left(\right. g \left.\right) + 4 H_{2} O \left(\right. l \left.\right)\) at constant temperature, ∆H – ∆E is:

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$\times {10}^5$ $\mathrm{Pa}$