Consider a hypothetical reaction between A, B, and C
that is firs...

Question

Consider a hypothetical reaction between A, B, and C
that is first order in A, zero order in B, and second order
in C. (e) By what factor does the
rate change when the concentrations of all three reactants
are tripled?

Accepted Answer

Welcome back everyone in this example, we're imagining a reaction with reactant X, Y, and Z. So we're told that the reaction is going to be second order with regards to X. And what that translates is that we have a coefficient of two in front of a reactant. X in our reaction. So then according to the prompt, we also have our reactant Y, which is a first order reactant. And so we would just have a coefficient of one in front of why? Whereas for our third reactant Z, we're told that we have a zero order respect of our reaction to the reactant. C. And so this would be our reaction here, but we're not given products. So we'll just place a question mark there. We're told that the concentration of all of our reactant is going to be doubled. And we determine which statement below is true about the rate change of our reaction. And because the statement is asking us to explain the rate change, we want to go ahead and write out a relationship for rate law for a given reaction. And we should recall that our rate law for any reaction is going to equal the rate constant K. Which we are not given a value for here, which is then multiplied by the concentration of each of our reactant raised to their coefficients as orders. And so for our first reactant we have the concentration of X, which according to the prompt is second order or is a second order reactant. And so we would say we have a concentration of X raised to the second power as an exponent here because it has a coefficient of two because it is a second order reactant. And so moving on to the second reacting here, we have the concentration of our reactant. Why? Which according to the prompt is a first order reactant and also has a coefficient of one in our given equation. And so or rather in our written equation. And so that is why we place an exponent of one here. And then lastly, we have our concentration of our third reactant. Z, which according to the prompt, is a zero order reactant. And so that means we place an exponents of zero in our rate law. And so now we want to go ahead and simplify this rate law. So we would simplify and say that our rate law is now equal to the rate constant K. Times the concentration of our reactant, X squared multiplied by the concentration of our second reactant, Y. Which is just a first order reaction. And so we simplified this exponent of one. Just the bracket here because it's implied. Now we should recall that for our third reactant, the concentration of Z raised to the zero order or zero power. We should recall that anything raised to the zero power is equal to one. And so we would just go ahead and plug in one here. And so to finally simplify our rate law, we would say that our rate law is equal to the rate constant K. Multiplied by the concentration of our reactant X. To the second order, or second power here, multiplied by the concentration of our second reactant Y, which is first order. And then we don't have to include that one there because we just understand that it is implied here. And so this would be our simplified rate law based on our given explained reaction. And so now we want to go ahead and recall that according to the prompt, the concentration of our reactant is going to be doubled. And so what this means is that this is going to change our rate law so that now when the concentration it's doubled, we would write out our rate law so that it states that our rate law is equal to our rate constant K. Times our concentration of our reactant X, which is now being doubled in concentration. So we would say two times our concentration of X, which is second order. So it still has that expletive to and then multiplied by our concentration of our reactant Y. Which is also doubled in volume now. So we would say or sorry, doubled in concentration now. So we would say two times why raised to the order one, which we just simplified here as just our bracket with Y. And so now that we have this being applied to each of our concentration of our reactant. We want to simplify this rate law so that we say that our rate law is equal to the rate constant K. Times our concentration of reacting X squared times the concentration of reactant Y. To the first order. Where we would recognize that two squared gives us four multiplied by two from our concentration of why being doubled. And so we have four times two which we can understand is simplified to eight. And so we can rewrite our rate law to say that our rate law changes by a factor of eight. So we would say our rate constant K. Is multiplied by the concentration of reacting X, which is second order times the concentration of reactant Y, which is first order. And sorry, this is a two here Multiplied by a value of eight. And so we we can see that we went from this simplified rate law to now doubling our concentration to give us this new simplified rate law. So this would be our final simplified rate law once our concentration of our reactant are doubled. And so what this means is that we went to such a higher factor here being eight, meaning that the only correct and true choice to complete this example is Choice B, which states that the rate increases by a factor of eight. And so B would be our final answer to complete this example. So I hope that everything that I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

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