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Given equation and a couple rates, how do you determine reaction order?
7. Given:
2A(g) + B(g) --> C(g) + D(g)
When [A] = [B] = 0.10 M, the rate is 2.0 M*s-1; for [A] = [B] = 0.20 M, the rate is 8.0 M*s-1; and for [A] = 0.10 M, [B] = 0.20 M, the rate is 2.0 M*s-1. The order of the reaction is :
I know that the answer will be 2, but I don't know how to get there. Help?
2 Answers
- HPVLv 71 decade agoFavorite Answer
Construct a chart of [A], [B], and rate
Trial No. . . . .[A] . . . .[B] . . . .rate (M/s)
. .1 . . . . . .0.10 . . . .0.10 . . . .2.0
. .2 . . . . . .0.20 . . . .0.20 . . . .8.0
. .3 . . . . . .0.10 . . . .0.20 . . . .2.0
To find the effect of changing [A] alone, look at Trials 3 and 2. [A] is double from 0.10 to 0.20 with [B] held constant. The rate changes by a factor of four. This is a 2nd-order relationship since 2^2 = 4. If we had tripled [A], the rate would have increased by a factor of 9 (3^2 = 9).
Now to B. In Trials 1 and 3, we doubled [B] from 0.10 to 0.20 while holding [A] constant. Note that there is NO CHANGE in the rate at 2.0 M/s. This is a zer-order relationship, when we change the concentration of a reactant but the rate stays the same.
The overall order of the reaction = order A + order B = 2 + 0 = 2.
- ?Lv 45 years ago
How can the value regulation = ok[C]. besides the reality that i don't comprehend the respond, I disagree. The sluggish step (fee determining) is: X + C --> Y this means that the value regulation is = ok[X][C] however, considering that [X] isn't a reactant/product, it could't be secure interior the value regulation. yet whilst [X] is expressed in terms of [A] and [B], then this is secure interior the value regulation equation. i'm uncertain a thank you to precise [X] in [A] and [B]!!!! i think of you're attempting to assert that simply by fact the formation of [X] is so rapid this is disregarded because it is going to pile up (have an extremely super concentration) to attend to react with [C]. hence the plenty smaller concentration of [C] would be certain the value. That is clever to a pair quantity. however, what in case you talk the reaction between X and C independently. concentration of [X] could maximum in all risk be secure interior the value regulation equation. Now, placed this form, the concentration of [A] and [B] at the instant are not that vast to furnish this very super (countless) concentration of [X]. And concentration of [C] is fairly super. in accordance on your thought the value regulation could then be fee = ok[X] and not fee = ok[C]. the value regulation is neither ok[X] or ok[C]. It could desire to be fee = ok[C][X] to take up attention the two reactants. X could desire to then be expressed in [A]and[B]. My question is, how do you show [X] in terms of [A][B]?
