Many physical and chemical processes are reversible. A reversible process is said to be in dynamic equilibrium when the forward and reverse processes occur at the same rate, resulting in no observable change in the system. Once dynamic equilibrium is established, the concentrations or partial pressures of all species involved in the process remain constant. Created by Jay.
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- Why would the reaction be at equilibrium in 0.2M and 0.6M? Shouldn't it be at equilibrium in 0.25M and 0.5M, because we need the number of moles of X2 be twice the number of 2X?(2 votes)
- Equilibrium means that the forward and reverse reaction rates are equal, not that the product and reactant concentrations be equal. We can have wildly different concentrations for both and a reaction can be at equilibrium. At equilibrium the reaction reaches a point where the concentrations of the chemicals do not appreciably change anymore because products are being formed at the same rate reactants are.
Hope that helps.(4 votes)
- Are X_2 and X actual gases? If so, which gases are they?(1 vote)
- So those aren't any specific gases. Using X2 and X is just a general way to write the equation which could use any elements we wanted. Think of X being used here as a variable in the same way that it's used in algebra.
Hope that helps.(2 votes)
- [Instructor] To illustrate the concept of equilibrium, let's say that we have a beaker and we put some water into our beaker. And also we make sure that our beaker has a lid on it. Some of those water molecules are going to evaporate and turn into a gas. And eventually once we have enough gaseous water, some of the gaseous water is going to condense and turn back into liquid water. To represent these two processes, we can show liquid water on the left and gaseous water on the right. So in the forward process, liquid water turns into gaseous water. And this forward arrow here represents the process of vaporization. And when gaseous water turns back into liquid water, that's represented by this arrow here on the bottom, and so that's the process of condensation. Since we start with liquid water, at first, the rate of vaporization is greater than the rate of condensation. But eventually we reach a point where the rate of vaporization is equal to the rate of condensation. And when that happens, if you're turning liquid water into gaseous water at the same rate, you're turning gaseous water back into liquid water, the number of water molecules in a liquid and gaseous state would remain constant. So when the rate of vaporization is equal to the rate of condensation, we've reached a state of equilibrium. And this is a dynamic equilibrium because if we zoom in and we look at this, water molecules are being converted from the liquid state to the gaseous state all the time and molecules are going from the gaseous state back to the liquid state all the time. However, since the rates are equal, the number of molecules in the liquid and gaseous state remained constant. And if we look at it from a macroscopic point of view, the level of water wouldn't change at all. Now let's apply this concept of dynamic equilibrium to a hypothetical chemical reaction. In our hypothetical reaction X2 which is a reddish brown gas, decomposes into its individual atoms to form 2X and individual atoms are colorless. So in the forward reaction, we're going from X2 to 2X. So X2 is decomposing to 2X. And in the reverse reaction, the two atoms of X are combining together to form X2. When we have a forward reaction and a reverse reaction by convention, we say, what's on the left side, are the reactants, and what's on the right side, are the products. And by using these terms, we can avoid confusion. Let's say that we start our reaction with only reactants. So only X2 is present in this first container here. And there are five particles of X2. If every particle represents 0.1 moles, since we have five particles of X2, we have 0.5 moles of X2. And let's say, this is a one liter container. 0.5 divided by one would be 0.5 molar. So the initial concentration of gas, X2 is 0.5 molar. And since we don't have any of the X, there are no white dots in this box, right? The initial concentration of X would be zero molar. So let me just write that in here, 0M. Next we wait 10 seconds. So we start off at time's equal to zero seconds, and now we're at time's equal to 10 seconds. And now we can see, there are three particles of X2 in our box. And so that would be 0.3M, so let's go ahead and write 0.3M in for our concentration. And now we have some particles of X. There are one, two, three, four particles. And once again, if each particle represents 0.1 moles that's 0.4 moles of X divided by one or 0.4M. We wait another 10 seconds, so when time is equal to 20 seconds, now there are two particles of X2 and one, two, three, four, five, six particles of X. So now the concentrations are 0.2M for X2 and 0.6M for X. We wait another 10 seconds for a total of 30 seconds. And there are still two particles of X2 and six particles of X. And so the concentrations after 30 seconds, the concentration of X2 is 0.2M and of X is 0.6M. Notice how the concentration of X2 went from 0.5M to 0.3M to 0.2, and then it was also 0.2 after 30 seconds. So it became constant when time is equal to 20 seconds. The concentration of X went from zero to 0.4 to 0.6 and then it was also 0.6 after 30 seconds. So the concentrations became constant when time is equal to 20 seconds, which means the reaction reached equilibrium after 20 seconds. So at time is equal to zero was not at equilibrium. When time is equal to 10 seconds, it was not at equilibrium. Only when time was equal to 20 seconds, did it reach equilibrium. And that equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. And if that's true, then X2 is being turned into 2X at the same rate that 2X is being turned back into X2. And if those rates are equal, the concentrations of X2 and X at equilibrium would remain constant. We can see the same concept, if you look at a graph of concentration versus time. Concentration of X2 starts at 0.5M when time is equal to zero seconds, and then drops to 0.3M after 10 seconds. And after 20 seconds, it's at 0.2M and then stays constant after that. For the concentration of X, we start out at 0M, we increase 2.4 after 10 seconds, then we're at 0.6 and then we are constant. So if you think about a line, if we just draw a dash line here at 20 seconds. That's the dividing line between on the left, where we're not at equilibrium. And so the concentrations are always changing. And then to the right of that dotted line, we are at equilibrium where the concentrations remain constant. So the equilibrium concentration of X2 gas is equal to 0.2M and the equilibrium concentration of X gas is equal to 0.6M. Finally, let's use these particular diagrams to think about what we would see at a macroscopic level as the reaction proceeds to equilibrium. And the first particulate diagram we see only red particles. So only our reactants X2 are present in the beginning. However, as time goes on, the number of red particles decreases from five in the first particular diagram to three in the second to two and then the number stays at two because remember we reach equilibrium after 20 seconds. So what we would see from a macroscopic point of view, is we'd start out with a reaction vessel that is a darker brown red, and then it would be a lighter brown red. And then finally, when we reach equilibrium and even lighter brown red, and it would stay that same light brown red, because we've reached equilibrium and the concentrations of reactants and products remain constant at equilibrium. Even though our reactants are turning into our products. Our products are turning back into our reactants at the same rate and therefore the concentrations of both reactants and products are constant.