The 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):

Given that the standard heats of combustion (ΔH°) of C(s), H₂(g), and CH₄(g) are −94 kcal/mol, −68 kcal/mol, and −213 kcal/mol, respectively.
What is the standard enthalpy change (ΔH°) for the reaction:

C(s) + 2H₂(g) → CH₄(g)?

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

The enthalpy of \(CH_4 + \dfrac{1}{2 }O_2 \rightarrow\ CH_3OH\)$$ is negative. If enthalpy of combustion of methane and  \(CH_3OH\) are x and y, respectively, then which of the following relations is correct?

The difference between enthalpy (∆H) and internal energy (∆E) for the given below reaction,
under constant temperature, is:
\(C_{3} H_{8} \left(\right. g \left.\right) + 5 O_{2} \left(\right. g \left.\right) \rightarrow 3 CO_{2} \left(\right. g \left.\right) + 4 H_{2} O \left(\right. l \left.\right)\)

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)