Answer:
0.0495 M/s
Explanation:
Rate of reaction (r) = Δ[[tex]CH_{3}CHO[/tex]]/Δt
Therefore, using the data given:
[tex]r_{1}[/tex] = (0.2480-0.1172)M/(1000-0)s = [tex]1.308*10^{-4}[/tex] M/s
[tex]r_{2}[/tex] = (0.1172-0.0767)M/(2000-1000)s = [tex]4.05*10^{-5}[/tex] M/s
Using the rate law:
[tex]r = k*[CH_{3}CHO] ^{m}[/tex]
r is the rate of the reaction (M/s), k is the rate constant (M/s), and m is a number.
Therefore, we have:
[tex]r_{1} = k*[0.1172]^{m}[/tex] (1)
[tex]r_{2} = k*[0.0767]^{m}[/tex] (2)
Divide equation (1) by equation 2, we have:
[tex]\frac{r_{1} }{r_{2} } = [\frac{0.1172}{0.0767}] ^{m}[/tex]
using [tex]r_{1} = 1.308*10^{-4} M/s[/tex] and [tex]r_{2} = 4.05*10^{-5} M/s[/tex]
We have:
[tex]\frac{1.308*10^{-4} }{4.05*10^{-5} } = [\frac{0.1172}{0.0767} ]^{m}[/tex]
Thus:
3.2296 = [tex]1.528^{m}[/tex]
Taking log of both sides, we have
log (3.2296) = m*log (1.528), and m = 2.77 (approximately 3)
Therefore, using equation (1) to get the rate constant (k), we have:
[tex]1.308*10^{-4} (M/s) = k*[0.1172]^{2.77}[/tex]
Thus k = 0.0001308/0.00264 = 0.0495 M/s
Thus, the rate constant is 0.0495 M/s
