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Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

Densities of ...

Question

Densities of diamond and graphite are $3.5$ and $2.3 g / m L$ respectively. Increase of pressure on the equilibrium:

$C_{(d i a m o n d)} ⇌ C_{(g r a p h i t e)}$

favours backward reaction

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favours forward reaction

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have no effect

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increases the reaction rate

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Solution

The correct option is A favours backward reaction
According to Le Chatlier's principle, if the pressure increases at equilibrium the reaction occurs in the direction where the volume decreases which is nothing but the density increased side.

As the density of the diamond is more than the graphite, the backward direction is favourable.

Hence option A is correct.

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Similar questions

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

and

2.3 g / m L

. Favourable conditions for formation of diamond are :

Q. Densities of diamond and graphite are

3.5

g/mL and

2.3

g/mL.

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:

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)}

and

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

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 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)}

to diamond

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

reduces its volume by

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

. If

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. Calculate the enthalpy change accompaying the conversion of 10 grams of graphite into diamond if the standard enthalpy of combustion of

C_{g r a p h i t e}

and

C_{d i a m o n d}

are -94.05 and -94.50 kcal respectively.

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Densities of diamond and graphite are 3.5 and 2.3g/mL respectively. Increase of pressure on the equilibrium:C(diamond)⇌C(graphite)

Densities of diamond and graphite are $3.5$ and $2.3 g / m L$ respectively. Increase of pressure on the equilibrium:

$C_{(d i a m o n d)} ⇌ C_{(g r a p h i t e)}$