For the reaction Cyclopropane$\rightarrow$propene A plot of ln[cyclopropane] vs time in seconds gives a straight line with slope -4.1 * 10^-³ s^-¹ at 550^$\omicron$C.What is the rate constant for this reaction? A) 3.9 * 10^-² s^-¹ B) 8.2 * 10^-³ s^-¹ C) 4.1 * 10^-³ s^-¹ D) 1.8 * 10^-³ s^-¹ E) 2.1 * 10^-³ s^-¹

Question

Quizplus · Accepted Answer

The rate law for this reaction can be written as: Rate = k[cyclopropane] Taking the natural logarithm of both sides, we get: ln(Rate) = ln(k[cyclopropane]) Using the properties of logarithms, we can rewrite this as: ln(Rate) = ln(k) + ln[cyclopropane] This is in the form of a linear equation, y = mx + b, where y = ln(Rate), x = time, m = rate constant k, and b = ln(k). From the given plot of ln[cyclopropane] vs time, we can determine the slope of the line, which is equal to the rate constant k. The slope is given as: slope = 4.1 F1F1F1S1F1F1F10 10F1F1F1S1^F1F1F10³ sF1F1F1S1^F1F1F10¹ Therefore, the correct answer is choice C: 4.1 F1F1F1S1F1F1F10 10F1F1F1S1^F1F1F10³ sF1F1F1S1^F1F1F10¹.

For the Reaction Cyclopropane →\rightarrow→ Propene a Plot of Ln[cyclopropane] Vs Time in Seconds Gives

For the Reaction
Cyclopropane $\rightarrow$ Propene
a Plot of Ln[cyclopropane] Vs Time in Seconds Gives