The kinetics of this reaction were studied as a function of tempe...

Question

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.

Accepted Answer

hi everyone for this problem we're told the table below shows the results of a kinetic study for the alkaline hydrogel icis of ethyl acetate. The reaction is second order overall calculate the value of the rate constant at 10 degrees Celsius. So our goal here is to calculate the value of our rate constant at 10 degrees Celsius. To solve this problem, we need to plot the given data and from there we can find the value of our activation energy and frequency factor and then use that to find the value of K. At 10 degrees Celsius. So one thing we're going to need to recall here is the Iranian equate and log form which is Ln of K is equal to negative activation energy over r universal gas constant times one over r temperature in kelvin units plus Ln of a. What this translates to is Y equals M X plus B. So when we plot this data, we're going to get an equation of the line made using this raw data and this is something you can plot either an excel or on your calculator. And when we find out what that equation is, it's going to be why equals negative 6876 X plus 20.997. Okay, so from this this equation that we derived, we have the value of the slope which is M and that is represented by negative 6876. And we have the value for the intercept B, which is 20.997. So we just need to find the value of X. Which is one over T four, So let's go ahead and do that. So we know that our temperature is 10 degrees Celsius but we need this in kelvin, so to go from degrees Celsius to kelvin, we add 273.15 and this is going to give us 283.15 kelvin. So now we can solve for X, which is one over T. So one over T is going to equal one over a temperature that we just converted to Kelvin, which is 283.15. So our one over T is going to equal 0.0035317. So now we can use our equation to solve for K. So why? Which is Ln of K is going to equal M. X plus B. Okay, so we have everything, we have em we have X and we have B. So we're going to solve for K. Which is what we're asked to calculate here the value of the rate constant. So we're going to get L n f k is equal to our slope M which is negative 6876 times the X. We just solved for or the X. Which is our one over T. It is 0. plus B. And we know what our B value was. So it's 20.997. So this simplified gives us L n f k is equal to negative 3.287. So remember we're solving for K here. So to get rid of that, L N. We're going to raise both sides to E. So que is going to equal E raised to negative 3.287. And when we do that we get a final answer of K is equal to 3.737 times to the negative two. And our units for K is leaders times mole inverse, times second inverse. So this is going to be our final answer for the Value of the rate constant at 10°C. That's the end of this problem. I hope this was helpful.

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