Rate Determination and Activation Energ DATA TABLE Rate oorstant, k (8-1) Trial
ID: 711873 • Letter: R
Question
Rate Determination and Activation Energ DATA TABLE Rate oorstant, k (8-1) Trial Tenm Cc) 122 . DATA ANALYSIS 1. Plot a graph of your data above, using Temperature (C) as the x-axis, and the rate constant, k as the y-axis 2. Determine the activation energy, E, by plotting the natural log of k vs. the reciprocal of absolute temperature. Calculate the activation energy, E, for the reaction. To do this, first calculate the best fit line equation for the data in Step 2. Use the slope, m, of the linear fit to calculate the activation energy, Ea, in units of kJ/mol. Note: On a plot of In k vs. 1/absolute temperature, E-mxR 3. A well-known approximation in chemistry states that the rate of a reaction often doubles for every 10°C increase in temperature. Use your data to test this rule. (Note: It is not necessarily equal to 2.00; this is just an approximate value, and depends on the activation energy for the reaction.) 4. Using the rate constant and precise temperature value for the trial that was done at room temperature (-20°C), as well as the Es value you obtained in Step 3 above, calculate the valu 5. the rate constant at 40°C. TmpInk T. -4.45 L .05 45 13055 -41.612 ,062/ -4.324 Advanced Chemistry with VernierExplanation / Answer
the plot of temperature ( deg.c) vs Rate constant (K) is plotted and shown below.
Arhenius equation is used to determine the rate constant for the reaction. The equation is
K= A*e(-E/RT) (1)
A = Frequency factor and E= activation energy and R= gas constant= 8.314 J/mole.K a nd T is temperature in K
taking ln, Eq.1 becomes lnK= lnA-E/RT
so a plot of lnK vs 1/T gives straight line whose slope is -E/R and intercept is lnA.
so a plot of lnK vs 1/T is shown below.
from the plot, slope -E/R= -4667 and E= 4667*8.314 J/mole =38801.44 J/mole
let the T1= 25 deg.c =25+273= 298K, T2= T1+10= 35 deg.c= 35+273= 308K
from Arhenius equation, ln(K2/K1)= 4667*(1/298-1/308)
K2/K1= 1.66, so very close to 2, but tihs ratio also depends upon the activation energy.
the equation of best fit is lnK= -4667/T+11.35
at t=40 deg.c= 40+273= 313K
lnK= -4667/313+11.35
K= 0.0284/sec
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