For the reaction cyclobutane(g) $\rightarrow$ 2ethylene(g)At 800 K,A plot of ln[cyclobutane] vs t gives a straight line with a slope of -1.6 s^-¹.Calculate the time needed for the concentration of cyclobutane to fall to 1/16 of its initial value. A) 2.3 s B) 1.7 s C) 1.3 s D) 0.63 s E) 1.6 s

Question

Quizplus · Accepted Answer

The slope of the plot of ln[cyclobutane] vs time is equal to the negative of the rate constant, k, for a first-order reaction. Given the slope as -0.693 s^-1, the time (t) required for the concentration to fall to 1/16 of its initial value can be calculated using the first-order kinetics formula: t = (ln(N0/N))/k, where N0/N = 16. Substituting the values, t = (ln(16))/0.693 = 1.7 s.

For the Reaction Cyclobutane(g) →\rightarrow→ 2ethylene(g)At 800 K,A Plot of Ln[cyclobutane] Vs T Gives a Ln[cyclobutane]

For the Reaction Cyclobutane(g) $\rightarrow$ 2ethylene(g)At 800 K,A Plot of Ln[cyclobutane] Vs T Gives a Ln[cyclobutane]