Answer to Question #79448 in General Chemistry for Chloe Lozano

Question

The rate of a standard reaction is 0.00543 M/s at 40 oC. What will the rate be if the temperature is doubled?

A. 0.01086 M/s
B. 0.02172 M/s
C. 0.04344 M/s
D. 0.08688 M/s
E. All of the Above

Answer 1

From the Arrhenius equation:
k = Ao exp(-Ea / RT)
for two conditions
k1 = Ao exp(-Ea / RT1)
k2 = Ao exp(-Ea / RT2)
dividing the second by the first
(k2 / k1) = exp( (-Ea/R) x (1/T2 - 1/T1) )
Since the general rate equation takes this form:
rate = k x [A]^n
if [A] is the same for 2 setups
rate1 = k1 x [A]^n
rate2 = k2 x [A]^n
dividing the second by the first
rate2 / rate1 = k2 / k1
so that
(rate2 / rate1) = exp( (-Ea/R) x (1/T2 - 1/T1) )
and finally
rate2 = rate1 x exp( (-Ea/R) x (1/(2xT1) - 1/T1) )
where
rate1 = 0.00543 M/s
Ea = ?
R = 8.314 J/molK
T1 = 40°C = 313.15K
T2 = ?

To solve this you need to know 2 things

(1) what 2 x T1 means. Is it 80°C or is it 626.3K... 2x40°C or 2x313.15K ? Does that "temperature doubled" refer to the °C temp or the absolute temp? The answer depends on this!
(2) what Ea is.. activation energy. Without out that, A, B, C, or D are all possible.

There is a sort of a rule of thumb that kind applies to some reactions at temps a bit above room temperature.
"for every 10°C rise in temp, the rxn rate doubles".
so... from 40°C to 80°C is a 4 x 10°C rise

rate1 ---> 2 x rate1 ---> 4 x rate1 --> 8 x rate1 ---> 16 x rate1
and the final rate would be 16 x 0.00543 = 0.08688 M/sec...
Answer: D.