Consider the reaction between Calcium Carbonate and Hydrochloric Acid. I understand the reaction rate can be increased by increasing the concentration of reactants. However I can't seem to understand the intuition behind it. If we were to increase the quantity of Hydrochloric Acid, would it not also increase volume of the reaction having no affect on the reaction rate. Furthermore does increasing both reactants increase the rate of the reaction? Increasing both would increase the volume meaning their would be the same concentration of Calcium Carbonate per unit volume of Hydrochloric Acid. So shouldn't the reaction rate not be affected.

  • 1
    $\begingroup$ Rates of heterogeneous reactions are always very tricky, as there is many unknown variables. $\endgroup$
    – Poutnik
    Nov 14, 2022 at 7:21
  • 2
    $\begingroup$ Quantity and concentration are different things. 2L of 36% HCl does not have higher concentration than 1L. But 1L of 36% HCl has higher concentration than 1L of 20% HCl. $\endgroup$
    – Poutnik
    Nov 14, 2022 at 7:23
  • 2
    $\begingroup$ Please try to use meaningful yet concise titles that reflect as much as possible what the question is about without opening it. Note that chemical names are common nouns and follow corresponding capitalization rules. $\endgroup$
    – andselisk
    Nov 14, 2022 at 7:52

1 Answer 1


Equations of reaction rates are based on statistical reasoning about collision frequency, which is supposed proportional to the reaction rate at given temperature.

Imagine $\pu{1 mmol/L}$ of $\ce{A}$ and $\pu{1 mmol/L}$ of $\pu{B}$ forms $\ce{C}$ at rate $\pu{1 mmol L-1 min-1}$ by a simple bimolecular reaction:

$$\ce{A(aq) + B(aq) -> C(aq)}$$

$\pu{2 mmol/L}$ of $\ce{A}$ and $\pu{1 mmol/L}$ of $\pu{B}$
would lead to doubled $\ce{A}$ and $\ce{B}$ collision frequency and doubled reaction rates.

$\pu{2 mmol/L}$ of $\ce{A}$ and $\pu{2 mmol/L}$ of $\pu{B}$
would lead to quadrupled $\ce{A}$ and $\ce{B}$ collision frequency and quadrupled reaction rates.


$$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} = k [\ce{A}][\ce{B}]$$

Rates of heterogeneous reactions are always very tricky, as there is many unknown variables. Such reactions of gases or liquids with solids complicate everything. As quantity more or less analogic to concentration, we could use the parameter:

$$\text{"surface density"} = \frac{\text{"total surface area of the solid"} }{\text{"(total) volume of suspension"}}$$

assuming suspension is evenly distributed in the volume. It would start failing for bigger chunks, sitting on the bottom, on at sedimentation of not well stirred mixture.

Note that this parameter has non trivial dependence of the reaction quotient (progress). It does not decrease linearly, as particle sizes decrease, particle surface/mass ratios increase and smaller particles react relatively faster.

If we imagine $\ce{B}$ is solid and reaction $\ce{A(aq) + B(s) -> C(aq)}$, it could be like:

$$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} = k [\ce{A}] \cdot \text{"surface density of B"}$$

But $k$ is not really a constant (for given $T$) here, as it depends to many factors, e.g. the way or intensity of mixing, or forming other phases lile a gas. There remain many unknown variables, mostly related to diffusion in heterogenous environment and surface layer phenomena. That makes models and rate equations reflecting reality semi-empirical at the best.


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