I got the first part but I have tried over and over again to get the last two parts of question correct but cannot. Please help, thanks.
A reaction X ---> 2 Y proceeds according to a second-order rate law, with k = 0.246 L mol-1min-1.
If [X]0 = 0.524 mol/L at t = 0:
A) What is [X] at t = 45 min?
B) Assuming that the initial concentration of Y is zero ( [Y]0 = 0 ) and that there is no reverse
reaction, what is the concentration of Y at 45 min?
C) What is the rate of the reaction at t = 45 min?
Answers:
A) 0.077 mol L^-1
B) 0.89 mol L^-1
C) 1.4E-3 mol L^-1 min^-1
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Verified answer
A)
1/[X] - 1/[X]o = kt
1/[X] - 1/0.524M = (0.246L/mol-min)(45 min)
[X] = 0.077M
B)
Δ[X] = [X] - [X]o
0.077 - 0.524M =
-0.447M
-Δ[X] = (+1/2)Δ[Y]
0.447M = (1/2)Δ[Y]
Δ[Y] = 0.89M
C)
rate = k[X]²
rate = (0.246L/mol-min)(0.077M)²
rate = 1.4e-3 M/L-min
the two the above thoughts are optimal acceptable aside from the sign distinctive than for the ensure in ok: on condition that -d[A]/dt = ok[A]^a million/2 (detrimental on condition that [A] is reducing), -ok dt = d[A]/[A]^a million/2 = [A]^-a million/2 d[A] and -ok t (t - 0) = 2[A]^a million/2 ([A] - [A0]) or -kt = 2[A]^a million/2 -2[A0]^a million/2. Rearranging, [A]^a million/2 = [A0]^a million/2 - (ok/2)t I all of unexpected met this subject attempting to derive the 2d order value regulation very final 3 hundred and sixty 5 days in AP chem (i'm a instructor), as straight away as I forgot the - sign in front of the differential.