Rate Law question

[Ashley3323]

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Hello everyone,
was wondering if anyone can explain this question ; the correct answer is D. Thank you in advance!

 

[Mna627]

well the NOCl2(g) cancels out because it is found in the product of the top reaction and as a reactant in the bottom reaction. Now, add the reactants from both reactions that were not canceled out and you have 2NO and Cl2. Rate will be k multiplied by the reactants so since you have 2NO and 1Cl2, the answer as you stated is choice D.

 

[510586]

well the NOCl2(g) cancels out because it is found in the product of the top reaction and as a reactant in the bottom reaction. Now, add the reactants from both reactions that were not canceled out and you have 2NO and Cl2. Rate will be k multiplied by the reactants so since you have 2NO and 1Cl2, the answer as you stated is choice D.

Oh yea so if the slow step is 2nd you add the ones above it, correct?

 

[Ashley3323]

well the NOCl2(g) cancels out because it is found in the product of the top reaction and as a reactant in the bottom reaction. Now, add the reactants from both reactions that were not canceled out and you have 2NO and Cl2. Rate will be k multiplied by the reactants so since you have 2NO and 1Cl2, the answer as you stated is choice D.

the part I don't understand is how cl2 becomes cl ?

 

[Lnguyen7]

the part I don't understand is how cl2 becomes cl ?

That gotta be a typo because

If you add two reactions, you will see NOCl2 is an intermediate.
2NO (g) + Cl2 (g) -> 2NOCl (g)
Now, you are likely to write a rate law based on this net reaction as Rate = k[NO]^2[Cl2] However, you cannot do that because the net reaction is not elementary. It is just a coincidence that the reaction order relative to NO and Cl2 are the same as their coefficient.

Since step 2 is RDS, the rate law will be: Rate = k2[NO][NOCl2]
However, we don't usually include intermediate in a rate law and so we have find [NOCl2] relative to the species that appear in the net equation NO and Cl2

We know from step 1
Rate forward = kf[NO][Cl2]
Rate reverse = kr[NOCl2]
Lets assume equilibrium is quickly reached so that rate f = rate r
Thus kf[NO][Cl2] = kr[NOCl2]
So [NOCl2] = (kf/kr)[NO][Cl2]

Substitute [NOCl2] into the rate law
Rate = k2[NO](kf/kr)[NO][Cl2]
Lets k2*kf/kr = k
Thus Rate = k[NO]^2[Cl2]

 

[Ashley3323]

That gotta be a typo because

If you add two reactions, you will see NOCl2 is an intermediate.
2NO (g) + Cl2 (g) -> 2NOCl (g)
Now, you are likely to write a rate law based on this net reaction as Rate = k[NO]^2[Cl2] However, you cannot do that because the net reaction is not elementary. It is just a coincidence that the reaction order relative to NO and Cl2 are the same as their coefficient.

Since step 2 is RDS, the rate law will be: Rate = k2[NO][NOCl2]
However, we don't usually include intermediate in a rate law and so we have find [NOCl2] relative to the species that appear in the net equation NO and Cl2

We know from step 1
Rate forward = kf[NO][Cl2]
Rate reverse = kr[NOCl2]
Lets assume equilibrium is quickly reached so that rate f = rate r
Thus kf[NO][Cl2] = kr[NOCl2]
So [NOCl2] = (kf/kr)[NO][Cl2]

Substitute [NOCl2] into the rate law
Rate = k2[NO](kf/kr)[NO][Cl2]
Lets k2*kf/kr = k
Thus Rate = k[NO]^2[Cl2]

that makes perfect sense now , thank you !!!

 

[510586]

That gotta be a typo because

If you add two reactions, you will see NOCl2 is an intermediate.
2NO (g) + Cl2 (g) -> 2NOCl (g)
Now, you are likely to write a rate law based on this net reaction as Rate = k[NO]^2[Cl2] However, you cannot do that because the net reaction is not elementary. It is just a coincidence that the reaction order relative to NO and Cl2 are the same as their coefficient.

Since step 2 is RDS, the rate law will be: Rate = k2[NO][NOCl2]
However, we don't usually include intermediate in a rate law and so we have find [NOCl2] relative to the species that appear in the net equation NO and Cl2

We know from step 1
Rate forward = kf[NO][Cl2]
Rate reverse = kr[NOCl2]
Lets assume equilibrium is quickly reached so that rate f = rate r
Thus kf[NO][Cl2] = kr[NOCl2]
So [NOCl2] = (kf/kr)[NO][Cl2]

Substitute [NOCl2] into the rate law
Rate = k2[NO](kf/kr)[NO][Cl2]
Lets k2*kf/kr = k
Thus Rate = k[NO]^2[Cl2]

Hey so I tried doing this with one of the examples Chad provided, and I am getting something very weird looking. Can you tell me if this strategy does not apply to the question or what I may be doing incorrectly? BTW treat the k's at the end as just one k like you said.

 

[SCDP]

Why wouldn't the answer be C? I thought based on Chad's the slow step is the rate determining so it would just be C.

If it's not C, that's stupid because in Chad's example:
O3 <---> O2 + O (fast)
O + O3 <---> 2O2 (slow)

rate = k[O][O3]

 

[510586]

Why wouldn't the answer be C? I thought based on Chad's the slow step is the rate determining so it would just be C.

If it's not C, that's stupid because in Chad's example:
O3 <---> O2 + O (fast)
O + O3 <---> 2O2 (slow)

rate = k[O][O3]

Same. Check the example I just posted it is that exact thingy haha 😛

 

[Lnguyen7]

Hey so I tried doing this with one of the examples Chad provided, and I am getting something very weird looking. Can you tell me if this strategy does not apply to the question or what I may be doing incorrectly? BTW treat the k's at the end as just one k like you said.

You are correct! You are going to use this strategy for problems like this which you are given several steps reaction.

 

[Lnguyen7]

Why wouldn't the answer be C? I thought based on Chad's the slow step is the rate determining so it would just be C.

If it's not C, that's stupid because in Chad's example:
O3 <---> O2 + O (fast)
O + O3 <---> 2O2 (slow)

rate = k[O][O3]

Rate law does base on the RDS, but when writing a rate law, we don't usually include intermediate, so we have to relate the concentration of the intermediate with something else

 

[510586]

@org

Rate law does base on the RDS, but when writing a rate law, we don't usually include intermediate, so we have to relate the concentration of the intermediate with something else

I guess Chad doesn't go in depth enough. I definitely remember doing this stuff in genchem. Is that notation OK or do you have to do [x]^-2? Not that it matters since it is multiple choice.