Relationship between forward and reverse reaction rates

relationship between forward and reverse reaction rates

If the rates of the forward and backward reaction are the same it means the reaction The 2nd part of your question is referring to stoichiometric relation of rates. They are equal, and NONzero. At dynamic chemical equilibrium, the rates of the forward and reverse reactions are equal to each other, i.e. Reactions at Equilibrium. As a reversible reaction proceeds, the forward rate slows down while the rate of the reverse reaction speeds up. Let's examine why this.

This actually makes sense if you think in terms of ideal gases. In such a mixture, each gas behaves as though it were the only one in the container. Adding an inert gas to an equilibrium has no effect on the concentrations or partial pressures of the gases in the reaction, so it has no effect on the rates of the forward or reverse reaction.

relationship between forward and reverse reaction rates

Because liquids and solids are highly incompressible, they do not change their volume and therefore they do not change their concentration when the size of the container or the external pressure changes. Thus, if there are no gaseous reactants or products, a change in pressure has no effect on the reaction rates. If both of these conditions are met, then reducing the volume squeezes the gaseous reactants and products into a smaller volume, increasing their concentration; while increasing the volume allows them to expand into a larger volume, decreasing their concentration.

Lesson 8: Chemical Reactions at Equilibrium and LeChatelier's Principle

Again, only reactants and products are affected. Thus, as you might suspect, the eventual effect of changing the volume depends on whether the gases are reactants, products, or both. Suppose, for example, that the only gases present are reactants. Decreasing the volume increases the reactant concentration, increasing the rate of the forward reaction, but it has no immediate effect on the rate of the reverse reaction because none of the product concentrations have changed.

Chemical Equilibrium Explained - Video Tutorial - Crash Chemistry Academy

Since none of the products are gases, decreasing the volume does not change their concentrations appreciably. Because the products are either solids or liquids they are already condensed and they cannot easily be compressed into a smaller volume.

This increase in the forward rate only causes a net loss of reactant, which slows the forward rate back down, and a net gain in the concentration of product, which increases the reverse rate, until the two rates are once again equal.

The shift to the right uses up the gaseous reactant. If the only gases present are products, then the reverse is true. Increasing the pressure by reducing the volume will cause a shift to the left.

CH Lesson 8 LeChatelier's Principle

If both reactants and products include gases, the direction of the shift will depend on the relative amounts. In this reaction, since there are more moles of gaseous reactants than gaseous products, reducing the volume will cause the equilibrium to shift to the right. In general, a change in pressure caused by a change in volume has a greater effect on reaction rate than if the number of moles of gas is greater. In this reaction, 4 moles of gas one mole of N2 and three moles of H2 become 2 moles of gas.

Shifting to the right reduces the amount of gas in the vessel, lowering the pressure, and countering the increase in pressure caused by reducing the volume.

But if we increase the pressure by adding some other gas, one not involved in the reaction, the equilibrium will not shift at all because neither the reactants nor the products will have changed concentration. This assumes, of course, that the other gas does not react with any of the chemicals that are involved in the reaction.

If it does, then that will cause a decrease in the concentration of the chemical the other gas reacts with, and this will cause a shift in the position of equilibrium.

relationship between forward and reverse reaction rates

For each reaction shown, decide which direction the equilibrium will shift when the pressure is increased by reducing the volume of the reaction vessel. If neither reactants nor products contain any gases, or if the number of moles of gaseous reactants is the same as the number of moles of gaseous products, changing the pressure by changing the volume will have no effect on the position of equilibrium.

You should have determined the following equilibrium shifts: There are 3 moles of gas on the reactant side and only two moles of gas on the product side. Shifting to the right will reduce the number of moles of gas, reducing the pressure — counteracting the increase in pressure that was imposed by the reduction in volume.

In this equation, X is the concentration of X at any moment in time, X 0 is the initial concentration of this reagent, k is the rate constant for the reaction, and t is the time since the reaction started. To illustrate the power of the integrated form of the rate law for a reaction, let's use this equation to calculate how long it would take for the 14C in a piece of charcoal to decay to half of its original concentration.

We will start by noting that 14C decays by first-order kinetics with a rate constant of 1. The integrated form of this rate law would be written as follows.

We are interested in the moment when the concentration of 14C in the charcoal is half of its initial value. Substituting this relationship into the integrated form of the rate law gives the following equation. We now simplify this equation and then solve for t. It therefore takes years for half of the 14C in the sample to decay.

relationship between forward and reverse reaction rates

This is called the half-life of 14C. In general, the half-life for a first-order kinetic process can be calculated from the rate constant as follows. Let's now turn to the rate law for a reaction that is second-order in a single reactant, X. The integrated form of the rate law for this reaction is written as follows.

Integrated form of the second-order rate law: