Does removing reaction products speed up the reaction rate?

The question should be whether it is the apparent reaction speed or the actual reaction speed.

For irreversible reactions.

Removing the product does not speed up the reaction.

Let's start with the reaction rate.

Definition. -ra) = k * [reactant order to the power of the reactant] that is, the reaction rate is only and the rate constant.

It is related to the concentration of reactants.

Let's look at the reversible reaction.

According to the above definition, assuming that equilibrium has been reached, the instantaneous removal of the product has no effect on the reactant concentration, so the instantaneous rate does not change.

However, with the extension of time, the equilibrium shifts positively, the concentration of reactants decreases, and the rate of positive reaction decreases.

This is the actual reaction rate.

But there is also the apparent reaction rate, due to the presence of the reverse reaction, the apparent reaction rate = positive reaction rate - hidden reaction rate.

The rate of the reverse reaction of the removed product decreases, so the apparent rate of reaction increases.

However, at the high school level, it is believed that removing the product only affects the balance, but not the rate.

No, but it can affect the equilibrium of the reaction and make the reaction move in the direction of the product.

Hello. Increase the concentration of substances (applicable to reactants, solids, and pure liquids) to accelerate the reaction rate (on negative and negative are increased by the same) The reverse rate increases with the original equilibrium state (the original inverse rate is equal) and the equilibrium changes at the moment when the conditions change. Le Chatre's principle considers: if the equilibrium system changes and affects the equilibrium conditions, the equilibrium must always be able to weaken the change of the species and increase the concentration of the concentration, like the reversal of the downward shift, the problem can be clarified.

Conditions that affect the reaction rate:

Temperature pressure. If the reactant of the catalyst is a gas, increasing the concentration of the reactant is equivalent to increasing the pressure, which will speed up the reaction rate, and if the product is a gas, removing the product is equivalent to decreasing the pressure, but will reduce the rate, and the solid and liquid are regarded as having no effect (theoretically, there is a little bit, the reason is also the pressure change, and the effect is very small and negligible).

Reaction rate increases The reaction rate slows down.

The first one does not speed up, the second speeds up.

The reaction rate should only be related to temperature, and the other thing that affects is the chemical equilibrium.

After the reversible reaction is equilibrated in the figure, the moment the product is removed, the v inversion decreases, and the v is constant. Subsequently, v is inversely increased and v is positively decreased, and a new equilibrium is established after the two are equal.

The positive reaction rate increases instantaneously, and the reverse reaction rate gradually increases, and finally equilibrates.

When some of the products (non-solid or non-pure liquids) are removed, the positive and reverse reaction rates are reduced, but the reverse reaction rate decreases relatively more. So the equilibrium moves in the positive direction, and you can follow the principle of Le Chatletle.

Determine the direction of balance movement.

As the equilibrium shifts in the positive direction, the concentration of reactants decreases; Although the positive movement of the balance increases the spawn, the increase is smaller than that of the initial move (as your teacher should have said: the equilibrium movement can change the change of the ascend, but it cannot counteract the change), so the result is that the concentration of the spawn also decreases.

Hope it helps!

For reversible reactions, when the chemical reaction begins, the rate of expression of reactants increases gradually, and the rate of the product increases gradually, due to the influence of solubility, temperature, etc. For the rate-time graph in the book, it means that in a reversible reaction, the reaction is theoretically considered to have reached equilibrium, the positive reaction has a maximum reaction rate, and the reverse reaction starts at 0, the positive reaction rate decreases, and the reverse reaction rate increases, until the positive and reverse reaction rates are equal.

The reactants decrease, and the products gradually increase. How could spawns also be reduced?