A researcher raises the temperature from 72.8 to 92.3 o C and finds that the rate of the reaction doubles. What was the activation energy (in J/mol) for this reaction? (R = 8.3145 J/mol K)

E = 37467J/mol = 37.5kJ

Explanation:

Please see attachment below.

The equation that relates two rates of reaction (Arrhenius equation) is what is used to solve this kinds of problem the equation is given as follows.

r1l2/r1= e^-E/R×(1/T2–1/T1)

Where

r1 = initial rate of reaction

r2 = final rate of reaction

E = Activation energy

T1, T2 = initial and final temperatures.

R = gas constant = 8.314J/milk

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j