The rate constant (k) for a reaction was measured as a function of temperature. a plot of lnk versus 1/t(in k) is linear and has a slope of −9.20×103 k . you may want to reference (pages 642 - 648) section 14.5 while completing this problem. part a calculate the activation energy for the reaction.

The Arrhenius equation is

so if we plot a graph between lnK (natural log of rate constant) and 1/T (inverse of temperature in kelvin)

The straight line equation will be

lnK = lnA - Ea/RT

Where

A = constant for a particular reaction at any temperature

R = gas constant = 8.314 J /molK

So as per this equation the slope of line will be = -Ea/R

Or

Ea = -slope X R

Ea = -(-9.20X10³X 8.314) =76.49kJ /mol

activation energy = 76.49 kJ /mol

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.