Answer:
Ea =22542.6
Explanation:
The rate constant k is affected by the temperature and this dependence may be represented by the Arrhenius equation:
[tex]k=Ae^{-\frac{E_a}{RT} }[/tex]
where the pre-exponential factor A is assumed to be independent of temperature, R is the gas constant, and T the temperature in K. Taking the natural logarithm of this equation gives:
ln k = ln A - Ea/(RT)
or
ln k = -Ea/(RT) + constant
or
ln k = -(Ea/R)(1/T) + constant
These equations indicate that the plot of ln k vs. 1/T is a straight line, with a slope of -Ea/R. These equations provide the basis for the experimental determination of Ea.
now applying the above equation in the problem
we can write that
[tex]ln\frac{k_2}{k_1} = \frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2} ][/tex]
solve for Ea:
Ea = R[Ln(k2/k1)] / [(1/T1) - (1/T2)]
but k_2 = 2 k_1, hence:
Ea = (8.314 J/moleK)[ln(2)] / [(1/273+45) - (1/273+73)]
Ea =22542.6
Answer:
The activation energy for this reaction is 22.6 kJ/ mol
Explanation:
Step 1: Data given
Rate constant doubles when Temperature goes from 45.0 °C to 73.0 °C
R = 8.314 J/K*mol
Step 2: Calculate the activation energy
Log (k2/k1) = Ea / 2.303R *((1/T1) - (1/T2))
⇒ with k1 = initial rate constant
⇒ with k2 = rate constant after doubled = 2k1
⇒ T1 = initial temperature = 45.0 °C = 318 Kelvin
⇒ T2 = Final temperature = 73.0 °C = 346 Kelvin
log (2) = Ea / (2.303*8.314) *((1/318) - (1/346))
log(2) = Ea / (2.303*8.314) * 0.00025448
Ea = 22649 J/mol = 22.6 kJ/mol
The activation energy for this reaction is 22.6 kJ/ mol
