The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.
Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)
Step 2: O(g) + O₃(g) → 2 O₂ (slow)
(b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.)

Question

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)

Step 2: O(g) + O₃(g) → 2 O₂ (slow)

(b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.)

Accepted Answer

welcome back everyone in this example, we have the reaction where hydrogen gas reacts with di atomic iodine gas to produce two moles of hydrogen iodide gas. This occurs in a two step mechanism. And we need to use this mechanism to determine the rate law for the given reaction. So what we should recognize is that we have a two step mechanism. And in step two we have that listed as a slow step. And so we should recall that our slow step is going to be our rate determining step. And so this is what we would base our final answer off of. And so because we know that this is the rate determining step, we can write out a rate law expression by saying that our rate for this step to rate determining step is equal to our rate constant for step two in our mechanism. So we would say K with a subscript of two, multiplied by the concentration of our reactant in step two raised to their coefficients. And so for our first reactant in step two we have our two moles of iodine gas. So we would say that K two, or the rate constant for step two is multiplied by the concentration of our first reactant, iodine. It has a coefficient of two. So we will raise that to the second power and then this is then going to be multiplied by the concentration of our second reactant being our hydrogen gas. So multiplied by H2, which has a coefficient of one in step two. And so we would raise it to a power of one. What these exponents also mean, is that gas is second order in respect to our reaction here, and hydrogen gas is first order with respect to our reaction. And so we want to simplify this rate law to say that our rate for our slow step is equal to our rate constant for step two, multiplied by the concentration of iodine gas to the second power and then multiplied by the concentration of hydrogen gas. Where we will just simplify the first power here as just a bracket here because it's implied and understood. And so this would be our simplified great expression for our rate determining slow step. But what we should recognize is that I dine is a unique reacting here because not only does it occur as a reactant in step to where it's consumed to form our products, but it also occurs as a product and step one in the fast equilibrium. And we recognize that we have this equilibrium arrow, meaning that our ID and gas is occurring not only as a product here, but technically also as a reactant, and that is due to the fact that I dine is acting as an intermediate here. So we would say that our two moles of iodine gas act as an intermediate. And that is due to the fact that in Step one, again, it's produced as a product, whereas in Step two, we recall that an intermediate should be consumed as a reactant and iodine gas does follow this action as an intermediate. And so because we know that it is in a fast equilibrium in step one, we can say that for our rate expression for step one we have our rate constant for step one, which is multiplied by the concentration of the reactant for the forward reaction here. And so that is just our di atomic iodine, So I two. And this should be set equal to because we have an equilibrium fast up here where we have our rate constant for the reverse reaction. Since this is an equilibrium times the concentration of our reactant for the reverse reaction, which will now be our two moles of iodine gas. And so we would say iodine and then we raise that coefficient of two as an exponent here. So this is specifically our equilibrium great expression that we have written out here and now that we have written out this equilibrium rate expression. And we know that our two moles of iodine gas here are intermediates or rather isn't intermediate. We can go ahead and for our rate determining rate law that we've written out for the slow step in Step two, we want to go ahead and find the concentration for our two moles of iodine. And so we want to take our rate expression from the equilibrium and step one and isolate for our concentration of our iodine to the second power. And so we'll just make some room down here to show how we find that. So in order to isolate for the concentration of ID and squared, we would divide both sides by our inverse or by our rate constant for the reverse reaction. So we divide both sides by K to the negative one subscript. And so in doing so we would have our concentration of our iodine which is now equal to because we can now cancel out the rate constant for the reversed reaction direction. We would have the rate constant for the forward reaction times the concentration of di atomic iodine divided by And let's make this divisor sign better. So divided by our rate constant for the reverse reaction. And so now that we have this expression where our concentration of two moles of iodine are isolated. Now for this expression here, we can take this expression and substituted in place of our rate determining rate law here for the concentration of iodine to the second power. So just to be clear we are substituting this four. The concentration of our I dine. And this is going to give us our rate law for a reaction. And so we can say that our rate law and I'll just scroll down for more room is now equal to or rather expressed as. So we'll just put a colon and will say that our rate. Sorry. So the rate is equal to the rate constants in the numerator where we would take our rate constant for the forward reaction and step one times the rate constant for step two. So we would have K one times K two. And then in our denominator we still have our inverse rate constant. Or rather the rate constant for the reverse reaction in step one. So we divide by that. So this expression is multiplied by our concentration of di atomic iodine. To complete our substitution of this entire expression for the concentration of Aydin to the second power here and now that we have substituted this expression here fully. We can now we're now left over with our concentration of H2 which is our other reactant. So we multiply this by our concentration of hydrogen gas. So now that we've outlined this expression, we can simplify this rate expression so that we can just express our rate constant instead of in three different terms. We just want to express it as just one term here because we know that it all will just equal one single value. And so we can say that our rate is equal to the rate constant. K. Multiplied by our concentration of di atomic iodine times the concentration of hydrogen gas. And so this statement here would be our final answer to complete this example as the rate law for the given mechanism. So I hope that everything that I explained was clear. If you have any questions, please leave them down below. And I will see everyone in the next practice video.

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.
Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)
Step 2: O(g) + O₃(g) → 2 O₂ (slow)
(b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.)

Verified Solution

Key Concepts

Reaction Mechanism

Rate Law

Elementary Steps

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.Step 1: O3(g) ⇌ O2(g) + O(g) (fast)Step 2: O(g) + O3(g) → 2 O2 (slow)(b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.)

Verified Solution

Key Concepts

Reaction Mechanism

Rate Law

Elementary Steps

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.
Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)
Step 2: O(g) + O₃(g) → 2 O₂ (slow)
(b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.)