The desorption (leaving of the surface) of a single molecular layer of n-butane from a single crystal of aluminum oxide is found to be first order with a rate constant of 0.128/s at 150 K. b. If the surface is initially completely covered with n-butane at 150 K, how long will it take for 25% of the molecules to desorb (leave the surface)? For 50% to desorb?

The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.)
C2H5Br(aq) + OH- (aq) → C2H5OH(l) + Br- (aq)
Temperature (°C) k (L,mol •s)
25 8.81⨉10-5
35 0.000285
45 0.000854
55 0.00239
65 0.00633
b. Determine the rate constant at 15 °C.
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Key Concepts
Reaction Order
Arrhenius Equation
Rate Constant (k)
This reaction has an activation energy of zero in the gas phase: CH3 + CH3 → C2H6
a. Would you expect the rate of this reaction to change very much with temperature?
Consider the two reactions:
O + N2 → NO + N Ea = 315 kJ/mol
Cl + H2 → HCl + H Ea = 23 kJ/mol
a. Why is the activation barrier for the first reaction so much higher than that for the second?
The evaporation of a 120-nm film of n-pentane from a single crystal of aluminum oxide is zero order with a rate constant of 1.92⨉1013 molecules/cm2•s at 120 K. a. If the initial surface coverage is 8.9⨉1016 molecules/cm2, how long will it take for one-half of the film to evaporate?
Consider the two reactions:
O + N2 → NO + N Ea = 315 kJ/mol
Cl + H2 → HCl + H Ea = 23 kJ/mol
b. The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, calculate the ratio of the reaction rate constants for these two reactions at 25 °C.
