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AP®︎/College Chemistry
Course: AP®︎/College Chemistry > Unit 7
Lesson 1: Introduction to equilibriumDirection of reversible reactions
For a reversible reaction, if the rate of the forward reaction is greater than the rate of the reverse reaction, there is a net conversion of reactants to products. If the rate of the forward reaction is less than the rate of the reverse reaction, there is a net conversion of products to reactants. If the two rates are equal, the reaction is at equilibrium. Created by Jay.
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How can we derive that relation (R = kf[X2] or R = kr[x]^1, here) ?(5 votes)
In the X₂ ⇌ 2X reaction, why are the forward and reverse reaction rates equal at equilibrium? From my understanding, dynamic equilibrium is when the number of reactant molecules being converted to product over time (the forward rate) equals the number of reactant molecules being created from product over time, making the number of reactant and product molecules remain constant. Thing is, the number of reactant molecules being created from product over time is not the same thing as the reverse rate (number of product molecules being converted to reactant over time), since it takes two product molecules to create one reactant molecule. Shouldn't the reverse reaction rate need to be higher to keep up with the two product molecules produced per reactant molecule by the forward reaction, to produce enough reactant to replace those converted to product?
(I know that rates are really change in concentration over time instead of change in particle count, but I can ignore that since the volume doesn't change, right? It is very possible my understanding of rates is flawed.)(2 votes)
At equilibrium, the rate of the forward and reverse reactions are equal. This is because equilibrium occurs when both the forward and reverse reactions are equal. For example, water(in liquid form) is at equilibrium when the rate of vaporization is equal to rate of condensation.(1 vote)
How will we know if a reaction is elementary? Will we be told and if, not, how can we determine if the reaction is indeed elementary?(2 votes)- Hi I noticed that when doing rates of reactions, it was said that we need to look at the experimental data to determine exponents on the rate law as these are not the same as the coefficients on the balanced equation. However on the example given here I saw something different.(1 vote)
Video transcript
- [Instructor] As an example
of a reversible reaction, let's look at the hypothetical reaction where diatomic gas X2 turns
into its individual atoms X, and it would turn into two of them. So X2 goes to 2X. The forward reaction
is X2 turning into 2X. And the reverse reaction
is 2X combining to form X2. And let's say the X2
is a reddish brown gas. If we assume that both the
forward and the reverse reactions are elementary reactions, we can actually write the rate law from the balanced equation. So for the forward reaction, let's go ahead and write the
rate of the forward reaction is equal to the rate constant
for the forward reaction, which we will symbolize
as K with the subscript F, times the concentration of, if we're going in the forward direction, the reactants would be X2. So times the concentration of X2. And since we have a coefficient
of 1 in front of X2, for this elementary reaction, this would be raised to the first power. Next we can write the rate
law for the reverse reaction. So the rate of the reverse reaction is equal to the rate constant, and we'll put in a subscript R here. So that's the rate constant
for the reverse reaction. And in the reverse reaction,
2X combines to form X2. So this would be times
the concentration of X. And since we have a
two as our coefficient, we need to raise the concentration
of X to the second power. Next, let's look at these
particulate diagrams and think about what happens
for the forward reaction. So we start at time is equal to zero, and we start with only X2. So here are five particles of X2. If we wait 10 seconds, now we've gone from five particles of X2 to only three particles of X2. So overall, two of those particles of X2 have turned into X. And so there are four particles of X in this second particular diagram. We wait another 10 seconds for a total of time is
equal to 20 seconds. And we've gone from three particles of X2 to only two particles of X2, and we've increased in the particles of X. So now we're up to six particles of X. So the concentration of X2 has decreased. We went from five particles
of X2 to three particles of X2 to only two particles of X2. And if we look at the rate
law for the forward reaction, the rate of the forward
reaction is proportional to the concentration of X2. So if the concentration of X2 decreases, the rate of the forward
reaction also decreases. We can see the same concept if we look at a graph of rate versus time. So if we look at this line right here, we're starting on a certain
rate for the forward reaction. And as the concentration of X2 decreases, we can see the rate of
the reaction decrease. So the rate of the
reaction stops decreasing when we get to time is
equal to 20 seconds. Next, let's think about the
rate of the reverse reaction. Well, when time is equal to zero, the rate of the reverse reaction is zero. And that's because when we
start out, we have only X2, we don't have any X present. So the reverse reaction doesn't happen. But as soon as some of
that X2 turns into X, it's possible for the
reverse reaction to happen. And as we increase in the amount of X, and we look at our rate law
here for the reverse reaction, as we increase in the concentration of X, the rate of the reverse reaction
should increase as well. And so that's why we see, that's why we see the rate of
the reverse reaction increase as time increases. So as the forward reaction is happening, the reverse reaction is also
occurring at the same time. However, we don't really see that when we look at our particular diagram. In our particular diagrams, we see a net conversion of X2 into 2X, for example, looking from
the first particular diagram to the second, we see that two particles of X2 have turned into four particles of X. And going from the second
diagram to the third diagram, we see that another particle of X2 has turned into 2X. And therefore we have six particles of X at time is equal to 20 seconds. So since we see a net conversion
of reactants to products in our particular diagram, the rate of the forward
reaction must be greater than the rate of the reverse reaction. And we can see that. So before time is equal
to 20 seconds here, if we look at our rates, let's just pick, for example, time is equal to 10 seconds, for the forward reaction, there's a higher rate than
for the reverse reaction. So at times equal to 20 seconds, the rate of the forward reaction becomes equal to the rate
of the reverse reaction. So here's the line on our graph, where the rates become equal, and also notice the rates
become constant at this point. And when the rate of the forward reaction is equal to the rate of
the reverse reaction, the reaction has reached equilibrium. So to the right of the dotted line, the reaction is at equilibrium. And to the left of the dotted line, the reaction is not at equilibrium. And since the rate of the forward reaction is equal to the rate of the
reverse reaction at equilibrium, X2 is turning into 2X at the same rate that 2X
is turning back into X2. Therefore, the concentrations of X2 and X at equilibrium remain constant. And we can see that when we
look at the particular diagrams where time is equal to 20 seconds, and time is equal to 30 seconds. So both of these particular
diagrams have two X2 particles and six X particles. Let's look at a summary of
what the rates of the forward and reverse reaction mean in terms of reactants and products. If the rate of the forward reaction is greater than the rate
of the reverse reaction, that means there's a net conversion of reactants to products. So therefore over time, the amount of reactants would decrease and the amount of products would increase. Eventually, the rate
of the forward reaction becomes equal to the rate
of the reverse reaction. And that means the
reaction is at equilibrium and there's no net change in the amounts of reactants or products. So reactants are turning into
products at the same rate the products are turning
back into reactants. And then finally, if the
rate of the reverse reaction is greater than the rate
of the forward reaction, there's a net conversion
of products to reactants. So products are turning into reactants faster than reactants are
turning into products. And the example that we looked at, the rate of the forward reaction was greater than the rate
of the reverse reaction. And eventually the rates became equal and the reaction reached equilibrium. If we had looked at an example
of this third case here where the rate of the
reverse reaction is greater than the rate of the forward reaction, eventually the two
rates would become equal and this reaction would
reach equilibrium too.