The second-order rate constant has been measured at different temperatures for the decomposition of nitrous oxide...

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Answer 1

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from above data that

now we have to calculate the ln(k) and 1/T (remembering to convert temperature to kelvin in the process)

bu using above table we can plot the graph between ln ln(k) and 1/T

-6.28 .00115
-4.48 .00108
-2.87 .00103
-1.41 .000978

-2 -3 -4 y = 26.943-29006 x Slope=-E/R -5 -6 -7 0.001 0.00105 0.0011 0.00115 -1 1/T (Kl)

we get a linear graph decreasing ln(k) with increasing 1/T. The slope of this line is equal to -activation energy/R where R is the gas constant.

Using my calculator I found the slope to be -29006

from graph -Ea/R

-29006=-activation energy/R, where R=8.314J/K*mol

Activation energy=2.41*10^5 J/mol

Activation energy=2.41*10^2 kJ/mol

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