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Introduction to Le Chatelier's Principle

Densities of ...

Question

Densities of diamond and graphite are $3.5$ and $2.3$ $g / m L$ respectively. Increase of pressure on the equilibrium:
$C (d i a m o n d) ⇌ C (g r a p h i t e)$

A

favours backward reaction

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B

favours forward reaction

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C

has no effect

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D

increases the reaction rate

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Solution

The correct option is A favours backward reaction
The given reaction is :-
$C (d i a m o n d) ⇌ C (g r a p h i t e)$

Given, Density of diamond $= 3.5 g / m L$
density of graphite $= 2.3 g / m L$

$∵$ density of diamond > density of graphite

As the pressure is increased on the system at equilibrium then by Le Chatelier's principle, the reaction will shift towards the formation of a denser product. So, the backward reaction is favoured.

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Similar questions

Q. Densities of diamond and graphite are

3.5

2.3

C (d i a m o n d) ⇌ C (g r a p h i t e)

Δ_{r} H = - 1.9

kJ/mole
Favourable conditions for formation of diamond are:

C (d i a m o n d) ⇌ C (g r a p h i t e)

Δ_{r} = - 1.9 k J / m o l e

Densities of diamond and graphite are

3.5

2.3 g / m L

. Favourable conditions for formation of diamond are :

Q. Calculate the enthalpy change accompaning the conversion of

10 g

of graphite into diamond if the heats of combustion of

C (g r a p h i t e)

C (d i a m o n d)

- 94.05 k c a l

- 94.50 k c a l

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 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)}

T = 298 K

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 b a r

, and substance should be pure at a given temperature. The conversion of graphite

C_{(g r a p h i t e)}

C_{(d i a m o n d)}

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)}

[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]

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