Question

INSTRUCTIONS: 1) Show gll work; 2) include units in calculation steps and final answer. Attach graph only. Additional attachments will not be graded 1. The kinetic data below was obtained for the reaction: Ae BAnswer the questions that follows In[A] 1/A] (atml) Time (s) Pressure A (atm) 0 2,000 5,000 8,000 12,000 15,000 502 335 180 95.5 41.7 22.4 a) [6 Points] Complete the missing data in the table. Plot the following graphs in MS Excel: i) In(A] vs t, and li 1/IA] vs t. Print and attach the 2 graphs as one page. b) [2 Points] Is the reaction first order or second order with respect to A? Explain. c) [2 Points] What is the magnitude of the rate constant? d) [2 Points] What is the value of half-life for the reaction? Show all work e) [3 Points] What is the pressure of A after 10,000 s?

Answer 1

Answer:

(1) The table is shown below

Time (s)	[A](atm)	ln[A]	1/[A]
0	502	6.21860012	0.001992032
2000	335	5.814130532	0.002985075
5000	180	5.192956851	0.005555556
8000	95.5	4.559126247	0.010471204
12000	41.7	3.730501129	0.023980815
15000	22.4	3.109060959	0.044642857

(a) The plote ln[A] vs time and 1/[A] vs time are shown below

(b) From the above plots, the plot ln[A] vs time gives a straight line with R2=0.99998.

Therefore the order of the reaction is first order with respect to A.

The plot 1/[A] vs time is not a best fit, therefore it is not a second order reaction. (1/[A]=kt+1/[A]0)

(c) The rate law is ln[A]=-kt+ln[A]0

y=mx+c

Therefore slope=rate constant

-k=-2.07754x10^-4 s^-1

k=2.07754x10^-4 s^-1.

(d) Half life for first order reaction is

t_1/2=0.693/k=0.693/(2.07754x10^-4 s^-1)

t_1/2=3335.67 s.

(e) Given t=10000s, the [A]=?

The equation of line is

ln[A]=(2.07754x10^-4 s^-1) t + 6.22501

ln[A]=(2.07754x10^-4 s^-1)10000 s + 6.22501

ln[A]=4.147

[A]=e^4.147=63.274 atm

Therefore pressure of A at 10000 s =63.274 atm.

Thanks and I hope you like it.