At elevated temperatures, methylisonitrile (CH₃NC) isomerizes to acetonitrile (CH₃CN) : CH₃NC (g) → CH₃CN (g)
The reaction is first order in methylisonitrile. The attached graph shows data for the reaction obtained at 198.9 °C. The rate constant for the reaction is ________ s^-1.

Question

At elevated temperatures, methylisonitrile (CH₃NC) isomerizes to acetonitrile (CH₃CN) : CH₃NC (g) → CH₃CN (g)
The reaction is first order in methylisonitrile. The attached graph shows data for the reaction obtained at 198.9 °C.

At elevated temperatures, methylisonitrile (CH<sub>3</sub>NC) isomerizes to acetonitrile (CH<sub>3</sub>CN) : CH<sub>3</sub>NC (g) → CH<sub>3</sub>CN (g) The reaction is first order in methylisonitrile. The attached graph shows data for the reaction obtained at 198.9 °C. The rate constant for the reaction is ________ s<sup>-1</sup>. A) -1.9 × 10<sup>4</sup> B) +1.9 × 10<sup>4</sup> C) -5.2 × 10<sup>-5</sup> D) +5.2 × 10<sup>-5</sup> E) +6.2

The rate constant for the reaction is ________ s^-1.

Quizplus · Accepted Answer

From the graph, we can see that the half-life of the reaction is about 200 minutes (the time it takes for the concentration of methylisonitrile to decrease by half). Since this is a first-order reaction, we can use the equation: ln([CH₃NC]t/[CH₃NC]0) = -kt where [CH₃NC]t is the concentration of methylisonitrile at time t, [CH₃NC]0 is the initial concentration, k is the rate constant, and t is time. Using the data from the graph, we can choose two points on the curve and use them to calculate the rate constant: ln(2.00/4.00) = -k(200) k = 5.2 x 10^-5 s^-1 So the answer is D, +5.2 x 10^-5 s^-1.

At Elevated Temperatures, Methylisonitrile (CH3NC)isomerizes to Acetonitrile (CH3CN): CH3NC (G)

At Elevated Temperatures, Methylisonitrile (CH₃NC)isomerizes to Acetonitrile (CH₃CN): CH₃NC (G)