Effect of surface area of particles on the rate of a reaction

For a reaction where one of the reactants is a solid, e.g.

2H_3O^+(aq)+CaCO_3(s)\rightarrow Ca^{2+}(aq)+CO_2(g)+3H_2O(l)

an increase in surface area of the solid increases the rate of the reaction. This is because the collision of hydroxonium ions with the carbonate is limited by the surface area of the latter.

As illustrated by the diagram above, a big cube of CaCO3 has six surfaces with a total area of 24 cm3 available for the hydroxonium ions to collide. However, if the cube is broken down into smaller cubes, the total surface area accessible by the hydroxonium ions is now 48 cm3. This leads to a higher frequency of effective collision and therefore, a higher rate of reaction.


The mass of a reacting mixture consisting of excess HCl and 2 samples of CaCO3 is monitored over time (see graph above).

Which of the curves is for:

i) 50 g of CaCO3 chips; ii) 50 g of finely grounded CaCO3?

When the reaction is repeated with 50 g of CaCO3 chips and H2SO4, the reaction stops with the bulk of the CaCO3 unreacted. Why?


Curve II corresponds to the finely grounded CaCO3 with a greater total surface area for reaction and hence a higher rate of reaction. The higher rate of reaction is deduced from the steeper gradient of curve II at (albeit a negative one) versus that of curve I.

The reaction stops prematurely when H2SO4 is used, due to the formation of insoluble CaSO4, which coats the surfaces of the unreacted chips, preventing further reaction.


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