We’re given the plot of ln k (y) vs. 1/T (x).
This means we need to use the two-point form of the Arrhenius Equation:
k = rate constant
Ea = activation energy (in J/mol)
R = gas constant (8.314 J/mol • K)
T = temperature (in K)
A = Arrhenius constant or frequency factor
This is also in the form of:
m = slope
b = y-intercept
m = -1.10x104 K
b = 33.5
t = 25°C + 273.15 = 298.15 K
(CH3)3CBr + OH- → (CH3)3COH + Br-
in a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH-. In several experiments, the rate constant k was determined at different temperatures. A plot of ln(k) versus 1/T was constructed resulting in a straight line with a slope value of -1.10 x 104 K and y-intercept of 33.5. Assume k has units of s-1.
c. Calculate the value of k at 25°C.
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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.