Answer to Question #121845 in General Chemistry for Tiffany

Question

The data below were collected for the following reaction at 500 ∘C:
CH3CN(g)→CH3NC(g)
Time (h) [CH3CN] (M)
0.0 1.000
5.0 0.794
10.0 0.631
15.0 0.501
20.0 0.398
25.0 0.316

Determine the value of the rate constant at this temperature.
k= ?hr-1

Answer 1

Answer

The concentration of CH₃CN when t = 0 h is 1.000 M

The concentration of CH₃CN when t = 5 h is 0.794 M

The concentration of CH₃CN when t = 10 h is 0.631 M

The concentration of CH₃CN when t = 15 h is 0.501 M

The concentration of CH₃CN when t = 20h is 0.398 M

The concentration of CH₃CN when t = 25 h is 0.316 M

So, log [CH₃CN] is 0 when t = 0 h

log [CH₃CN] is -0.100 when t = 5 h

log [CH₃CN] is -0.199 when t = 10 h

log [CH₃CN] is -0.301 when t = 15 h

log [CH₃CN] is -0.400 when t = 20 h

log [CH₃CN] is -0.500 when t = 25 h

If we draw a graph between log [CH₃CN] and time (h), it will be a straight line passing through origin. So, the reaction is first order with respect to CH₃CN. The value of value of slope will be the reaction rate. For a first order reaction having rate constant k, the value of slope = -(k/2.303)

The rate of the reaction can be determined in an alternative way  i.e.

(0-(-.100))/(0-5) = ((-.100)-(-.199))/(5-10) = ((-.199)-(-.301))/(10-15) = ((-.301)-(-.400))/(15-20) = ((-.400)-(-.500))/(20-25) = - 0.01 h^-1

The reaction is first order with respect to CH₃CN

So, -(k/2.303) = - 0.01 h^-1

K = 0.02303 h^-1 (Answer)