Question

*****Help me find Ea in j/moles with the slope of -5056.2!!

Experiment 4 CHM 112 8. Use the following data to prepare an Arrhenius plot to determine activation energy. Follow the excel directions in the procedure section of this lab. a. Use the concentrations from mixture 1 in question 1 to calculate the rate constant, k b. Complete the rest of the table. c. in excel, plot Ink) vs 1/Temp. Use the graph to determine the slope and then the activation energy E. Remember, R is a constant, R 8.314 In(k) - RT+ constant +273 Temp., °C Temp., K Ice bath 277 Cold water 8 281 Room temp. 24 2017 Hot water 1307 Time Rate 445 , 2472 307b. 252 145 .2.2012 61 llo39154 1/Temp. Ln(k) 1/K 1.03001083 0.003557769 Dic3320034 1200323232 Copy and paste your excel plot here: In(k) v-5056.2x+37.957 1119.6766 2) 20.0480 3) 20.7980 4) 21.6637 20.5 19.5 0.0032 0.0033 0.0034 0.0035 0.0036 00037 2.2472=k(802). CK = 0.3511x08 g k (colds & 3.2573=k coo2Y:008).033 Slape -SOSTE 2 (K = 0.5090x108 kroom Coop766 7002 003) (02 -5056,2=-59/83145 = 1,0776808 78.34s (hot) 4019 r 16,39341=$1.002) (1008) (.02 K: 2.59158/09 Ea _ joules/mole

Answer 1

The Arrhenius equation is

La Ink =) + constant

The equation of straight line is

Y = mx +0

Slope

-E m =-

Eq= -Rx m

E = -8.314 X -5056.2 K-1)

EQ = 42037 mol

Hence, the activation energy is

42037 mol