Consider the reaction A+2BC whose rate at 25 C was measured using three differen
ID: 877569 • Letter: C
Question
Consider the reaction
A+2BC
whose rate at 25 C was measured using three different sets of initial concentrations as listed in the following table:
7.7×103
What is the rate law for this reaction?
Express the rate law symbolically in terms of k, [A], and [B].
Part A
where R is the gas constant (8.314 J/molK)
The activation energy of a certain reaction is 38.2 kJ/mol . At 26 C , the rate constant is 0.0120s1. At what temperature in degrees Celsius would this reaction go twice as fast?
Express your answer with the appropriate units.
SubmitHintsMy AnswersGive UpReview Part
Incorrect; Try Again; 3 attempts remaining
Part B
Given that the initial rate constant is 0.0120s1 at an initial temperature of 26 C , what would the rate constant be at a temperature of 200 C for the same reaction described in Part A?
Express your answer with the appropriate units.
Trial [A](M) [B]
(M) Rate
(M/s) 1 0.40 0.010 1.9×103 2 0.40 0.020 3.8×103 3 0.80 0.010
7.7×103
Explanation / Answer
1)
A + 2B <-------------------> C
rate= k [A]x [B]y
1.9 x 10^-3 = k [0.4]x [0.010]y -----------------> (1)
3.8 x 10^-3 = k [0.4]x [0.020]y ------------------> (2)
7.7 x 10^-3 = k [0.8]x [0.010]y ------------------> (3)
if we solve 1 diveded by 2 we get y = 1, if we solve 3 diveded by 1 we get x = 2
x = 2 , y = 1
rate = k [A] 2 [B] 1
rate = k [A] 2 [B] 1
1.9 x 10^-3 = k [0.4] 2 [0.010] 1
k = 1.187 L2 mol2 s1
PART A :
log(k2/k1) = ( Ea/2.303R)[1/T1-1/T2]
log (2k1/k1) = (38.2 / 2.303 x 8.314 x 10^-3) [1/299-1/T2]
T2 = 313 K
T2 = 400C
PART B :
T1 = 26 + 273 = 299 K
T2 = 200 + 273 = 473 K
k1 = 0.0120 s^-1
log(k2/k1) = ( Ea/2.303R)[1/T1-1/T2]
log (k2/ 0.0120) = (38.2 / 2.303 x 8.314 x 10^-3) [1/299-1/473]
log (k2/ 0.0120) = 2.45
(k2/ 0.0120) = 281.83
k2 = 3.382 s-1
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.