Second-order reaction and integrated rate law

  • #1

Homework Statement


The second-order rate constant for the reaction A+2B --> 2C+D is 0.34 dm3/mol s. What is the concentration of C after 20 s and 15 min if the initial concentrations were [A] = 0.027 mol/dm3 and [ B] = 0.130 mol/dm3

Homework Equations


The integrated rate law for A+2B --> P
$$k_r t=\frac{1}{[B ]_0-2[A]_0}\ln\frac{[A]_0([B ]_0-2x)}{([A]_0-x)[B ]_0}$$
with $$x=[P]$$ as it is given in Physical Chemistry by Atkins.

The Attempt at a Solution


Solving the above equation for x gives
$$x=\frac{[ B](\exp(k_r t([ B]-2[A]))-1)}{\frac{[ B]}{[A]}\exp(k_r t([ B]-2[A]))-2}.$$
By inserting the numbers from the problem statement I get x = 0.014485 mol/dm3.
But I can't really figure out if [C] is just 2/3 *x, since we have 2C and 1D as the products or what the final answer is.

thanks
 

Answers and Replies

  • #2
DrClaude
Mentor
7,617
4,053
Imagine that the reaction proceeds through an intermediary:
$$
\mathrm{A} + 2 \mathrm{B} \rightarrow \mathrm{P} \rightarrow 2 \mathrm{C} + \mathrm{D}
$$
with the final step being extremely fast. What would you get then?
 
  • #3
Imagine that the reaction proceeds through an intermediary:
$$
\mathrm{A} + 2 \mathrm{B} \rightarrow \mathrm{P} \rightarrow 2 \mathrm{C} + \mathrm{D}
$$
with the final step being extremely fast. What would you get then?

I'm really not sure what you want me to realize by this. Then
$$-\frac{d[P]}{dt}= k' [P]^2$$
but I don't see how this helps me determine the concentration of [C].
 
  • #4
DrClaude
Mentor
7,617
4,053
You calculated (I haven't checked that number) that x = 0.014485 mol/dm3 = [P]. Consider that P instantaneously breaks down into 2C + D, what do you get for [C]?
 
  • #5
Well I have the instantaneous reaction rates
$$ -\frac{d[P]}{dt}=\frac{1}{2}\frac{d[C]}{dt}$$
which gives
$$k'[P]^2=\frac{1}{2}\frac{d[C]}{dt}\Rightarrow \int_0^t k[P]^2 dt = \frac{1}{2}\int_0^{[C]}d[C]\Rightarrow k[P]^2 t = \frac{[C]}{2}$$
but if it is instantaneous t --> 0 and the above is just [C]=0, which can't be true.. So I guess this is the wrong way to go..
Btw I denoted k' since I wasn't sure if it can be set equal to the first k. Should they be equal when P is considered to break down instantaneously?
 
  • #6
Borek
Mentor
28,813
3,315
Think about just stoichiometry, you are overcomplicating things.
 
  • #7
Think about just stoichiometry, you are overcomplicating things.
That was what I tried in the original post. I mean we have P --> 2C+D. So the concentration of C is 2/3 of the original concentration, that is [C]=2/3 * [P]?
 
  • #8
21,468
4,854
That was what I tried in the original post. I mean we have P --> 2C+D. So the concentration of C is 2/3 of the original concentration, that is [C]=2/3 * [P]?
Suppose I told you that, for every mole of P produced, 2 moles of C are formed, and one mole of D is formed. What would you think?

Chet
 
  • #9
Hm okay. So since 1 mol P is proportional to 2 moles C and we go from P to C the conversion factor is 2 mol C / 1 mol P, that is [C] = 2 [P] ?
 
  • #10
21,468
4,854
Hm okay. So since 1 mol P is proportional to 2 moles C and we go from P to C the conversion factor is 2 mol C / 1 mol P, that is [C] = 2 [P] ?
That's what the chemical reaction balance equation you wrote says. You didn't need me to tell you this.

Chet
 

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