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collected: For the reaction 3NON,O + NO the following data were co t (min) INO]

ID: 1045824 • Letter: C

Question

collected: For the reaction 3NON,O + NO the following data were co t (min) INO] InNO 1/INO 0.01 -4.60517 100 60 o.00683 4.98643 146.4129 120 o.00518 5.26295 193.0502 180 0.00418 -5.47744 239.2344 240 0.0035 -5.65499 285.7143 300 0.00301 -5.80582 332.2259 360 0.00264 -5.93698 378.7879 time time 6· The order of this reaction in NO is 7. What is the magnitude of the rate constant for the reaction? 8. The half-life of this reaction is approximately 9. The concentration NO at 100 minutes will be approximately . The reaction 2NOBr 2NO+Br2 exhibits the rate law Rate kNOBr]? ANOBr] where k = 1.0 x 10-5 Mfi s-1 at 25°C. This reaction is run where the initial concentration of NO (INOBr]o) is 0.18 M. What is one half-life for this experiment?

Explanation / Answer

Q6) From the given plots, it can be seen that the best fit linear curve is obtained for 1/[NO] vs time, which suggests that the reaction is second order with respect to NO.

Explanation:

Second order integrated rate equation is given by,

2NO--->product

1/[NO]=1/[NO]o+kt

where [NO]=concentration of species X after time t

[NO]o=concentration of species X after time t=0

k=rate constant

The equation of 1/[NO] vs t resembles linear equation of the form y=mx+c ,with slope=k, intercept=1/[NO]o

Q7) slope=k=rate constant=(y2-y1)/(x2-x1), where (y2,x2) and (y1,x1) are two points on the graph

(y2,x2)=(193.0502,120)

(y1,x1)=(146.4129,60)

slope=(193.0502-146.4129)/(120-60)=0.777 M^-1 min^-1

Q8) At half life,

[NO]=1/2[NO]o=0.5[NO}o

1/0.5[NO]o=1/[NO]o+kt

or,1/0.5[NO]o-1/[NO]o=kt

or, 1-0.5/[NO]o=kt

0.5/[NO]o=k*t(half)

1/[NO]o=100=y-intercept

0.5*100=(0.777 M^-1 min^-1)*t(half)

or, t(half)=(0.5*100)/(0.777 M^-1 min^-1)=64.350 min

half life=64.350 min

Q9)t=100 min

1/[NO]=1/[NO]o+kt=100 M^-1+(0.777 M^-1 min^-1)*100min

1/[NO]=177.7 M^-1

[NO]=1/(177.7)=0.00563 M