Courses for Kids
Free study material
Offline Centres
Store Icon

Densities of diamond and graphite are 3.5 and 2.3 g/ mL respectively. Increase of pressure on equilibrium on:
C(diamond) $\leftrightharpoons$ C(graphite)
A) favours backward reaction
B) favours forward reaction
C) has no effect
D) increases the reaction rate

Last updated date: 16th May 2024
Total views: 405.3k
Views today: 4.05k
405.3k+ views
Hint: We need to use Le Chatelier’s principle of the effect of change of pressure on equilibrium reaction. We know that pressure and volume are inversely proportional based on this Le Chatlier’s principle states that an increase in pressure or volume makes the equilibrium to move in the direction that decreases other quantities.

Complete step by step answer:
> We can convert Graphite to Diamond at very high pressure and temperatures. But we know that this conversion is a highly endothermic reaction which requires a lot of activation energy. And the conversion of diamond to graphite is far more difficult and activation energy is more than that of graphite to diamond reaction activation energy.
> We know that Diamond and Graphite equilibrium is obtained at very high pressures from there on if we increase pressure reaction shifts towards the low volume side (From Le Chatelier’s principle).

> Given that the density of graphite is 2.3g/mL this implies that 1gm graphite has a volume of $\dfrac { 1 }{ 2.3 }$mL volume.
> Given that the density of Diamond is 3.5g/mL this implies that 1gm diamond has a volume of $\dfrac { 1 }{ 3.5 }$ mL volume.
> In the Diamond and Graphite equilibrium amount converted from one form to another will be the same so the volume of the diamond will be less than that of graphite so equilibrium will move in a backward direction.

Therefore, option A is correct.

Note: We may think that both reactant and product are solids so pressure does not affect equilibrium but at very high pressures it even affects solids. We may also confuse with Volume and density and can commit a mistake so we need to be careful while reading questions and answering.
Recently Updated Pages