Question

The rate constant doubles when the temperature is increased from 45.0 C to 73.0 C. What is the activation energy for this reaction? (R=8.314 J/Kmol)?

Answer 1

The activation energy for this reaction is 22.6 kJ/ mol

Step-by-step 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

Answer 2

Ea =22542.6

Step-by-step explanation:

The rate constant k is affected by the temperature and this dependence may be represented by the Arrhenius equation:

`k=Ae^{-(E_a)/(RT) }`

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

`ln(k_2)/(k_1) = (E_a)/(R)[(1)/(T_1)-(1)/(T_2)  ]`

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