Question

rates of chemical reactions

Plz help!! how do I find specific rate constant k, 1/T, In K

thank you

Rates of Chemical Reactions: Report Form Rate Law Summary: (a) From your measurements, what are the reaction orders with respect to each of the three reactants H;O; HOO (b) What is the rate expression (rate law) for the uncatalyzed reaction? Rate = K LT-] [H202) (c) What is the total reaction order? 2 3. Effect of Temperature on the Reaction Rate and the Activation Energy, E. ) Calculate the specific rate constant at each temperature and complete the table below. toon temp. 20-29°c Reaction time (s) (C) T(K) specific rate 1/T (K Mixture constant k 130 rit rit 6 97 134.5 307.5 l 12.0 285 2.5 275.5 Use Excel to plot Ink vs. 1/T. (Attach Excel plot to your report form.} From your Excel plot enter the Trendline here: 8 ] 6x10 Determine A/S,0,?) - {0-15.00 mLX0.050 M)}- - 1.666x10 (150.0 mL) and calculate the reaction Rate - . 1 1/8,0, 1. 2 At and specific rate constant, ki = Rate / ["H2021" [H], where k=1, and 6-8. for reaction mixtures 1, and 6-8, below and enter result in table above. Show specific rate constant calculation for reaction mixture 1, k1, below and give the values of the other three. Also show calculation of E, below. ki - E.(kJ/mol) -

Answer 1

how do I find specific rate constant k, 1/T, In K

As per your data/information :

Rate law , rate = k [H2O2] [I-]

for Reaction mixture #1 :

d {[S2O32-]} = 0- 5 ml*0.050 M / 150 mL = - 0.00167 M
rate of reaction : -1/2 * d[S2O32-]/dt   = 0.00167 M /130s = 0.0000128 M/s

k [H2O2] [I-] = 0.0000128 M/s

[H2O2] = 0.8 M * 10 mL/150 ml = 0.0533 M     ; [I-] = 0.050 M * 25 mL/150 ml = 0.00833 M

or   k [H2O2] [I-] = 0.0000128 M/s            ====>   k = 0.0000128 M/s / ( 0.0533 M    * 0.00833 M )

thus, specific rate constant , k = 0.029 M-1 s-1.

== 1/T = 1/T (K)

Reaction mixture #1 : 1/298.15 K = 0.003354 K-1

=====

lnk : natural log of specific rate constant , k

Reaction mixture #1 : ln (k = 0.029 ) = -3.54