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Suppose I am reacting HCl with CuCO3. Experiment 1: I use 2g powdered CuCO3 Experiment 2: I use 4g powdered CuCO3 All other variables are kept constant. Will this have any affect on rate of reaction? Will the 4g react faster because of more particles available for collision?

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  • $\begingroup$ More powder doesn't mean higher rate but a more saturated solution does. $\endgroup$ – Nisarg Bhavsar May 9 at 8:30
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    $\begingroup$ Higher mass means high surface contact area and therefore higher rate, especially if HCl is in sufficient excess. $\endgroup$ – Poutnik May 9 at 8:43
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$1$) Let's admit first that the reaction rate is proportional to the amount of reactants, in the usual way. Whatever the order $1$ or $2$ of the reaction, the rate constant ${k_{1} }$ or ${k_2}$ will not change if the amount of reactant is increased. But the rate of reaction ${r = - dn/dt}$ does increase when the amount of reactant ${n}$ increases, whatever the order.

For $1$st order, the rate is : ${r = - dn/dt = k_1 n}$. So the rate ${r}$ increases by increasing the amount of reactant ${n}$.

For $2$nd order, the rate of rection is : ${r = - dn/dt = k_2 n^2}$. Here too, the rate of reaction increases when the amount of reactant $n$ increases.

$2)$ As the reactant is a solid, the reaction rate may not be proportional to the amount of reactant (in mole). Let's suppose that it is proportional to the surface of the grains. Even so, the surface increases by increasing the amount of powder. As a consequence, the reaction rate does also increase.

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