Question

The rate of a certain reaction was studied at various temperatures. The table shows temperature (?) and rate constant (?) data collected during the experiments. Plot the data to answer the questions. What is the value of the activation energy, ?a , for this reaction? ? (?) ? (?−1) (400, 0.0000173) (420, 0.000127) (440, 0.000773) (460, 0.00403) (480, 0.0183) (500, 0.0738) (520, 0.267) (540, 0.879) (560, 2.66) (580, 7.43) ?a= ______ . kJ⋅mol−1 What is the value of the pre‑exponential factor (sometimes called the frequency factor), ? , for this reaction? ?= _______s−1  

Answer 1

E_a = 139 kJ/mol

A = 2.45 x 10¹³ s^-1

Explanation

When we plot the graph of ln(k) vs 1/T, we get a straight line whose equation is : y = -16717x + 30.828

slope = -16717 K

Activation energy, E_a = -(slope) * R

where R = constant = 8.314 J/mol-K

E_a = -(-16717 K) * (8.314 J/mol-K)

E_a = 138985.1 J/mol

E_a = 139 kJ/mol

intercept = ln(A)

30.828 = ln(A)

A = e^30.828

A = 2.45 x 10¹³ s^-1