Chemical Kinetics and Rate Law Question

[betterfuture]

---

Members don't see this ad.

Can someone explain why the answer is A. The explanation says as the [C] of KI doubles so does O2. I don't see KI doubling in [C] and neither does O2 double in formation.

For example, taking taking trial 2 and 3, the concentration goes from 0.060 to 0.090 does not double. And the formation of O2 does not double (7.25e-8 ----> 1.08e-7 is not doubling)

Thanks!

 

[Gurby]

Compare trial 1 and 2: [H2O2] doubles, while [KI] remains constant. In this situation, rate of O2 formation doubles. Thus, we know reaction is 1st order with respect to H2O2.

Compare trial 1 and 3: [KI] increases by 50%, while [H2O2] stays constant. Rate of O2 increases by 50%. Thus, 1st order with respect to [KI].

 

[betterfuture]

Okay, this may seem like a stupid question, and it probably is, but is doubling the same thing as increasing by 50%?

 

[Gurby]

Okay, this may seem like a stupid question, and it probably is, but is doubling the same thing as increasing by 50%?

Doubling is increasing by 100%, or being multiplied by 2.
Increasing by 50% is like being multiplied by 1.5.
If there is no change, you multiply by 1.

 

[betterfuture]

It's been too long since I took any math. That makes sense. But I still don't understand how KI is first order kinetics.

The requirement for first order kinetics states that as you double the [C] of reactant → there is double rate of product formation. As you stated, KI does not double, it increased by 50% (1.5 x [KI]) so..? Am I missing something here? Doesn't that violate first order kinetics or what?

 

[Gurby]

They increased [KI] by 50% and the reaction rate increased by 50%. What do you think would happen if you increased [KI] by 100%?

 

[theonlytycrane]

@betterfuture For first-order kinetics you will see a reaction rate increase by the same factor as the increase in reactant concentration. The doubling rule is an easy example where the reaction rate doubles since you doubled the reactant concentration. For [KI], the reactant concentration was increased by a factor of 1.5, and keeping [H2O2] constant, the reaction rate is increased by the same factor of 1.5 supporting first order kinetics in terms of [KI].

 

[betterfuture]

Hmm. Well what about 2nd order kinetics? Would they throw some kind of curveball here?

I understand with 1st order kinetics you said, the same factor of reactant concentration is seen in the product formation. Regarding 2nd order kinetics, when reactant concentration doubles, the product formation quadruples, typically. But in the not so simple case of 2nd order kinetics, say the reactants did not double, they instead increased by some factor, let's say 1.5, how would the product formation be changed here?

 

[theonlytycrane]

@betterfuture It's based on the rate law. If, for example, the rate law was k[KI]^2[H2O2] then increasing the concentration of [KI] by a factor of 1.5 would increase the rate by a factor of 1.5^2.

 

[aldol16]

I understand with 1st order kinetics you said, the same factor of reactant concentration is seen in the product formation. Regarding 2nd order kinetics, when reactant concentration doubles, the product formation quadruples, typically. But in the not so simple case of 2nd order kinetics, say the reactants did not double, they instead increased by some factor, let's say 1.5, how would the product formation be changed here?

At the very minimum, you should be able to understand "proportional to the square of" or "inversely proportional to the square of" for the MCAT, because many physical laws (e.g. gravitation, electrostatic, etc.) operate under inverse square laws. So in this case, second order kinetics means dP/dt = k*A^2. Therefore, if A goes up by a factor of 1.5, you can just plug in. dP/dt = k*(1.5*A)^2 = k*(1.5)^2*A^2. This means that dP/dt must go up by a factor of 1.5^2.

 

[betterfuture]

Oh I see. I am so used to just seeing a pattern that I forgot that I can just plug the numbers in. Makes sense now.