Scenario 1 describes the reaction that takes place between H2 and I2 to form the acid HI. This reaction involves very intricate and steps that are critical to understanding the concept of equilibrium as a whole. First, an important concept to understand when describing any reaction is the idea of collision probability. When we examine gas molecules in a 1 cm3 container, there can be billions of collisions per second between neighboring molecules, and even then, maybe only one of these collisions result in a reaction. If any one collision occurs, the only way they will react to form a new molecule, in this case an HI molecule, is if they meet specific activation energy requirements. This means that each collision must have enough energy to “pay
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Along with this, we can expect that with time (say 15 minutes), an equilibrium between the two gases and the newly formed HI compound will be established. This occurs when the molar concentrations of both the reactants and the products are equal. This is not to say that the reaction is finished, instead, the reaction will continue occurring until the molecules decay (which could take hundreds of years). Theoretically, the beaker 2 minutes after equilibrium is reached will yield the about same concentrations of products and reactants as 20 minutes after equilibrium is reached. Even though there is no change in the molar concentration of the gases in beaker, the H2 and I2 gas molecules are still reacting, this is simply happening at the same rate that the HI compounds are splitting apart, also called the reverse reaction. It is understood that when the rate of the forward reaction and the rate of the reverse reaction are equal, an equilibrium is established causing no further net change in molar concentration of reactants or …show more content…
Using the same principals as the scenario above, we can make predictions as to what will occur in this new container. When the 2HI molecules collide, given the proper collision orientation and activation energy, we will see the 2HI molecule split apart, or dissociate, into its two components, H2 and I2. This will continue to occur until a similar equilibrium represented in scenario 1 is reached between the products and reactants. As the second scenario approaches equilibrium, we will see an increase in pressure in the system until equilibrium is reached, where it will level off and reach constant pressure. This occurs for the same reason as in scenario 1, except here we are splitting molecules apart causing a net increase in pressure. When we examine the rate of reaction for the two scenarios, we must look at the equilibrium constant for each reaction. The first reaction has an equilibrium constant represented by the equation Keq = [HI]2 / [I2][H2] (using the rule: products over reactants, coefficients become exponents), where the reverse reaction is represented by the equation Keq = [I2][H2] / [HI]2. Though both of these scenarios will reach a similar equilibrium, these two equations show that the rates of these reactions will be different,
Along with this, we can expect that with time (say 15 minutes), an equilibrium between the two gases and the newly formed HI compound will be established. This occurs when the molar concentrations of both the reactants and the products are equal. This is not to say that the reaction is finished, instead, the reaction will continue occurring until the molecules decay (which could take hundreds of years). Theoretically, the beaker 2 minutes after equilibrium is reached will yield the about same concentrations of products and reactants as 20 minutes after equilibrium is reached. Even though there is no change in the molar concentration of the gases in beaker, the H2 and I2 gas molecules are still reacting, this is simply happening at the same rate that the HI compounds are splitting apart, also called the reverse reaction. It is understood that when the rate of the forward reaction and the rate of the reverse reaction are equal, an equilibrium is established causing no further net change in molar concentration of reactants or …show more content…
Using the same principals as the scenario above, we can make predictions as to what will occur in this new container. When the 2HI molecules collide, given the proper collision orientation and activation energy, we will see the 2HI molecule split apart, or dissociate, into its two components, H2 and I2. This will continue to occur until a similar equilibrium represented in scenario 1 is reached between the products and reactants. As the second scenario approaches equilibrium, we will see an increase in pressure in the system until equilibrium is reached, where it will level off and reach constant pressure. This occurs for the same reason as in scenario 1, except here we are splitting molecules apart causing a net increase in pressure. When we examine the rate of reaction for the two scenarios, we must look at the equilibrium constant for each reaction. The first reaction has an equilibrium constant represented by the equation Keq = [HI]2 / [I2][H2] (using the rule: products over reactants, coefficients become exponents), where the reverse reaction is represented by the equation Keq = [I2][H2] / [HI]2. Though both of these scenarios will reach a similar equilibrium, these two equations show that the rates of these reactions will be different,