Calculate thange in enthalpy for a given reaction,

2H₂O₂(l)$\to$ 2H₂O(l) + O₂(g)

(Given the heat of formation of H₂O₂(l) and H₂O(l) are -188 and -286 kJ/mol
respectively)       

1. -196 kJ/mol             

2. + 196 kJ/mol

3. +948 kJ/mol             

4. -948 kJ/mol

One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K)$\to$(4.0 atm, 5.0 L, 245 K) with a change in internal energy, $∆$U = 30.0 L atm. The change in enthalpy ($∆$H) of the process in L atm is:

1. 40.0

2. 42.3

3. 44.0

4. not defined, because pressure is not constant

According to the law, a perfect crystal's entropy is 0 at absolute zero, which is:

1. The first law               

2. Second law

3. Third law             

4. None of the above

At constant pressure and temperature, the direction of any chemical reaction is one where the...... decrease.

1. Entropy               

2. Enthalpy

3. Gibbs energy       

4. None of the above

The work done by a mass less piston in causing an expansion $∆$V (at constant temperature), when the opposing pressure, P is variable, is given by:

1. W= $-\int \mathrm{P}∆\mathrm{V}$               

2. W=0

3. W= -P$∆$V                     

4. none of these

Entropy decreases during:

1. crystallization of sucrose from solution

2. rusting of iron

3. melting of ice

4. vaporization of camphor

Entropy change of vaporization at constant pressure is given by:

1. $∆{\mathrm{S}}_{\mathit{(p}\mathit{)}}= \frac{∆{\mathrm{Hp}}_{}}{\mathrm{T}}$

2. $∆{\mathrm{S}}_{\mathit{\left(}v\mathit{\right)}}=\frac{∆{\mathrm{U}}_v}{\mathrm{T}}$

3. $∆{\mathrm{S}}_{\mathit{\left(}v\mathit{\right)}}= \frac{∆{\mathrm{H}}_v}{\mathrm{P}}$

4. None of the above

What is the entropy change for the reaction given below, 

2H₂ (g) + O₂ (g) $\to$2H₂O(l)

at temperature 300 K? Standard entropies of H₂ (g), O₂(g) and H₂O(l) are 126.6, 201.20 and 68.0 JK^-1mol^-1 respectively.

1. -318.4 JK^-1mol^-1                           .

2. 318.4 JK^-1mol^-1 

3. 31.84 JK^-1mol^-1                             

4. none of these

If S$°$ for H₂, Cl₂ and HCl are 0.13, 0.22 and 0.19 kJ K^-1mol^-1 respectively. The total change in standard entropy for the reaction, H₂ + Cl₂ $\to$2HCl is:

1. 30 J K^-1mol^-1                         

2. 40 J K^-1mol^-1

3. 60 J K^-1mol^-1                         

4. 20 J K^-1mol^-1

The direct conversion of A to B is difficult and thus it is converted by path A$\to$C$\to$D$\to$B. Given

$∆{\mathrm{S}}_{\mathrm{A}\to \mathrm{C}}=50 \mathrm{e}.\mathrm{u}; ∆{\mathrm{S}}_{\mathrm{C}\to \mathrm{D}}=30 \mathrm{e}.\mathrm{u}; ∆{\mathrm{S}}_{\mathrm{B}\to \mathrm{D}}=20 \mathrm{e}.\mathrm{u}.$

(Where e.u. is the entropy unit)
then $∆{\mathrm{S}}_{\mathrm{A}\to \mathrm{B}}$ would be:

1. +60 e.u                                       

2. +100 e.u

3. -60 e.u                                       

4. -100 e.u