Here is the correct question:
The reaction A → products was studied at a series of different temperatures. A plot of ln(k) vs. 1/T gave a straight line relationship with a slope of -693 and a y-intercept of -0.425. Additionally, a study of the concentration of A with respect to time showed that only a plot of 1/[A] vs. time gave a straight line relationship. What is the initial rate of this reaction when [A] = 0.41 at 271 K ?
Answer:
the initial rate of this reaction is 0.0216275 M/sec
Explanation:
Using the formula:
[tex]K = Ae^{\frac{-Ea}{RT}}[/tex]
[tex]InK = InA + (\frac{-Ea}{R})(\frac{1}{T})\\y \ \ \ \ \ \ \ \ \ \ c \ \ \ \ \ \ \ \ \ \ m \ \ \ \ x[/tex]
[tex]m = \frac{-Ea}{R}[/tex] [tex]= -693[/tex]
[tex]\frac{Ea}{8.314}= 693 \ \\ \\ Ea = 693 * 8.314 \\ \\ Ea = 5671.602 \ J[/tex]
[tex]In A = -0.425 \ \ \\ \\ A = e^{-0.425} \\ \\ A = 0.6538[/tex]
[tex]K = 0.6538 e^{- (\frac{5761.602}{8.314*271})[/tex]
[tex]K = 0.05275 \\ \\ K = 5.275*10^{-2}[/tex]
Since [tex]\frac{1}{[A]}[/tex] vs time is a straight line relationship;
Therefore, it is a second order reaction
rate = K[A]²
rate = 5.275 × 10⁻² × (0.41)
rate = 0.0216275 M/sec
