Asked by tere
A flask is charged with 0.124 mol of A and allowed to react to form B according to the reaction A(g) → B(g). The following data are
obtained for (A) as the reaction proceeds:
Time (s) 1 10 20 30 40
Moles of A 0.124 0.110 0.088 0.073 0.054
1) The average rate of disappearance of A between 10 s and 20 s is __________ mol/s.
answer: 2.2 × 10-3
2) If the rate law for the reaction
2A + 3B → products
is second order in A and first order in B, then the rate law is rate = __________.
answer: k(A)2(B)
i know the answers but how do i figure problems like the above? i have a bunch of similar problems to figure out.
obtained for (A) as the reaction proceeds:
Time (s) 1 10 20 30 40
Moles of A 0.124 0.110 0.088 0.073 0.054
1) The average rate of disappearance of A between 10 s and 20 s is __________ mol/s.
answer: 2.2 × 10-3
2) If the rate law for the reaction
2A + 3B → products
is second order in A and first order in B, then the rate law is rate = __________.
answer: k(A)2(B)
i know the answers but how do i figure problems like the above? i have a bunch of similar problems to figure out.
Answers
Answered by
DrBob222
10 sec, 0.110
20 sec, 0.088
0.110-0.088 = 0.022 moles/10 sec = 0.0022 moles/sec.
rate constant = k[A]<sup>x</sup>[B]<sup>y</sup>
where x is the order of A and y is the order of B and [A] and [B] are the concentrations, in moles/L, of the reactants. Except for elementary reactions, the order CAN NOT be determined from the coefficients in the balanced equation. x and y must be determined experimentally.
20 sec, 0.088
0.110-0.088 = 0.022 moles/10 sec = 0.0022 moles/sec.
rate constant = k[A]<sup>x</sup>[B]<sup>y</sup>
where x is the order of A and y is the order of B and [A] and [B] are the concentrations, in moles/L, of the reactants. Except for elementary reactions, the order CAN NOT be determined from the coefficients in the balanced equation. x and y must be determined experimentally.
Answered by
tere
thank you i kept getting 0.022 because i didn't divide by time.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.