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Entropy

Calculate Δ...

Question

Calculate $Δ S_{r}^{o}$ at $298 K$ of
(a) $N a (s) + \frac{1}{2} {C l}_{2} (g) \to N a C l (s)$ , (ii) $\frac{1}{2} N_{2} (g) + 2 H_{2} (g) + \frac{1}{2} {C l}_{2} (g) \to {N H}_{4} C l (s)$
(iii) $C$ (graphite) $\to C$ (Diamond)
The values of $S^{o}$ of $N a, {C l}_{2}, N a C l, {N H}_{4} C l, N_{2}, H_{2}$ diamond and graphite are $51, 223, 72, 95, 192, 131, 2.43, 5, 69 J K^{- 1}$ and ${m o l}^{- 1}$ respectively:

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Solution

(i) $N a (s) + \frac{1}{2} C l_{2} (g) ⟶ N a C l$
$△ S = △ S_{p r o} - △ S_{r e a c} = 72 - 51 - 223 = - 202$ .
(ii) $△ S = 95 - 223 - 131 - 192 = - 451$
.(iii) $△ S = 2.43 - 5.69 = - 3.26$ .

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Q. The standard state Gibbs free energies of formation of

C (g r a p h i t e)

C (d i a m o n d)

at T = 298 K are

Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 k J m o l^{- 1}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by

2 \times 10^{- 6} m^{3} m o l^{- 1}

C (g r a p h i t e)

is converted to

C (d i a m o n d)

isothermally at T = 298 K, the pressure at which

C (g r a p h i t e)

is in equilibrium with

C (d i a m o n d)

is:
[Useful information:

1 J = 1 k g m^{2} s^{- 2}

1 P a = 1 k g m^{- 1} s^{- 2}

1 b a r = 10^{5} P a

]

Q. The standard state Gibbs free energies of formation of

C (g r a p h i t e)

C (d i a m o n d)

at T = 298 K are

Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 k J m o l^{- 1}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by

2 \times 10^{- 6} m^{3} m o l^{- 1}

C (g r a p h i t e)

is converted to

C (d i a m o n d)

isothermally at T = 298 K, the pressure at which

C (g r a p h i t e)

is in equilibrium with

C (d i a m o n d)

is:
[Useful information:

1 J = 1 k g m^{2} s^{- 2}

1 P a = 1 k g m^{- 1} s^{- 2}

1 b a r = 10^{5} P a

]

Q. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298K are

Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 K J m o l^{- 1}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by

2 \times 10^{- 6} m^{3} m o l^{- 1}

. If C(graphite) is converted to C(diamond) isothermally at T = 298K, the pressure at which C(graphite) is in equilibrium with C(diamond), is
[Useful information:

1 J = 1 k g m^{2} s^{- 2}

1 P a = 1 k g m^{- 1} s^{- 2}; 1 b a r = 10^{5} P a

Q. The enthalpy of transition of C(diamond) to C(graphite) is?
Given that the enthalpies of combustion of C(diamond) and C(graphite) are – 393.5 and – 395.4 kJ

m o l^{- 1}

C

\to

C

Δ S_{300} = 10

m o l^{- 1}

C

+ O_{2} \to C O_{2}

Δ H = - 91

m o l^{- 1}

300

K

C

+ O_{2} \to C O_{2}

Δ H = x

300

x

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