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

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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:

$K = Ae^{\frac{-Ea}{RT}}$

$InK = InA + (\frac{-Ea}{R})(\frac{1}{T})\\y \ \ \ \ \ \ \ \ \ \ c \ \ \ \ \ \ \ \ \ \ m \ \ \ \ x$

$m = \frac{-Ea}{R}$ $= -693$

$\frac{Ea}{8.314}= 693 \ \\ \\ Ea = 693 * 8.314 \\ \\ Ea = 5671.602 \ J$

$In A = -0.425 \ \ \\ \\ A = e^{-0.425} \\ \\ A = 0.6538$

$K = 0.6538 e^{- (\frac{5761.602}{8.314*271})$

$K = 0.05275 \\ \\ K = 5.275*10^{-2}$

Since $\frac{1}{[A]}$ 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