Question

The reaction 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 ln[A] vs. time gave a straight line relationship. What is the initial rate of this reaction when [A] = 0.41 at 271 K?

Answer 1

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

Step-by-step explanation:

Using the formula:

`K = Ae^{(-Ea)/(RT)}`

`InK = InA + ((-Ea)/(R))((1)/(T))\\y \ \ \ \ \ \ \ \ \ \ c \ \ \ \ \ \ \ \ \ \ m \ \ \ \ x`

`m = (-Ea)/(R)`
`= -693`

`(Ea)/(8.314)= 693 \ \\ \\ Ea = 693 * 8.314 \\ \\ Ea = 5671.602 \ J`

`In A = -0.425 \ \ \\ \\ A = e^(-0.425) \\ \\ A = 0.6538`

`K = 0.6538 e^{- ((5761.602)/(8.314*271))`

`K = 0.05275 \\ \\ K = 5.275*10^(-2)`

Since
`(1)/([A])` 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