At 500 K in the presence of a copper surface, ethanol decomposesaccording to the following equation.
C2H5OH(g) CH3CHO(g) + H2(g)
The pressure of C2H5OH was measured as afunction of time and the following data were obtained.
Time (s) PC2H5OH(torr)
0 250.
100. 237
200. 224
300. 211
400. 198
500. 185
Since the pressure of a gas is directly proportional to theconcentration of gas, we can express the rate law for a gaseousreaction in terms of partial pressures. Using the above data,deduce the rate law, the integrated rate law, and the value of therate constant, all in terms of pressure units in atm and time inseconds. Predict the pressure of C2H5OH after900. s from the start of the reaction. [Hint: To determine theorder of the reaction with respect to C2H5OH,compare how the pressure of C2H5OH decreaseswith each time listing.]
To determine the order of the reaction with respect to C2H5OH, we need to compare how the pressure of C2H5OH decreases with each time listing. Let's calculate the change in pressure for each time interval:
Change in pressure for the first 100 seconds = 237 torr - 250 torr = -13 torr
Change in pressure for the second 100 seconds = 224 torr - 237 torr = -13 torr
Change in pressure for the third 100 seconds = 211 torr - 224 torr = -13 torr
Change in pressure for the fourth 100 seconds = 198 torr - 211 torr = -13 torr
Change in pressure for the fifth 100 seconds = 185 torr - 198 torr = -13 torr
Since the change in pressure is the same for each time interval, the reaction is first order with respect to C2H5OH. This means the rate law can be expressed as:
Rate = k[C2H5OH]
To determine the rate constant (k), we can choose any set of data points and solve for k. Let's choose the first and second data points.
Using the first data point:
Rate = k[250 torr]
Rate = k(250 torr)
Using the second data point:
Rate = k[237 torr]
Rate = k(237 torr)
Since the rate is the same for both data points:
k(250 torr) = k(237 torr)
Dividing both sides by k:
250 torr = 237 torr
Therefore, the rate constant (k) is equal to 250 torr/237 torr = 1.0557.
Now that we have the rate constant, we can determine the integrated rate law. Since the reaction is first order, the integrated rate law can be expressed as:
ln([C2H5OH]t/[C2H5OH]0) = -kt
Plugging in the values, we get:
ln([C2H5OH]t/250) = -(1.0557 s^(-1))t
To predict the pressure of C2H5OH after 900 seconds, we can rearrange the integrated rate law and solve for [C2H5OH]t:
ln([C2H5OH]t/250) = -(1.0557 s^(-1))(900 s)
[C2H5OH]t/250 = e^(-1.0557 s^(-1))(900 s)
[C2H5OH]t = 250e^(-1.0557 s^(-1))(900 s)
Calculating this using a calculator, the pressure of C2H5OH after 900 seconds is approximately 66.28 torr.
To determine the rate law for the reaction, we need to compare how the pressure of C2H5OH changes with time and identify the order of the reaction with respect to C2H5OH.
Let's analyze how the pressure of C2H5OH decreases with each time listing:
From time 0s to 100s, the pressure decreases from 250 torr to 237 torr.
From time 100s to 200s, the pressure decreases from 237 torr to 224 torr.
From time 200s to 300s, the pressure decreases from 224 torr to 211 torr.
From time 300s to 400s, the pressure decreases from 211 torr to 198 torr.
From time 400s to 500s, the pressure decreases from 198 torr to 185 torr.
It is clear that the pressure of C2H5OH decreases by a constant amount with each time interval. This suggests that the reaction follows first-order kinetics with respect to C2H5OH.
The rate law for the reaction can be expressed as:
Rate = k[C2H5OH]^1
Next, let's determine the integrated rate law. The integrated rate law can be obtained by integrating the rate law expression and solving for the concentration of C2H5OH as a function of time.
The integrated rate law for a first-order reaction is given by:
ln([C2H5OH]t/[C2H5OH]0) = -kt
Where [C2H5OH]t is the concentration of C2H5OH at time t, [C2H5OH]0 is the initial concentration of C2H5OH, k is the rate constant, and t is time.
In this case, since the pressure of C2H5OH is proportional to its concentration, we can use the partial pressures instead of concentrations.
ln(Pt/P0) = -kt
Where Pt is the pressure of C2H5OH at time t, P0 is the initial pressure of C2H5OH, k is the rate constant, and t is time.
Now, let's use the given data to calculate the rate constant, k.
Using the data at time 0s, Pt = 250 torr and t = 0s, we have:
ln(250 torr/250 torr) = -k(0s)
ln(1) = 0
0 = 0
Using the data at time 500s, Pt = 185 torr and t = 500s, we have:
ln(185 torr/250 torr) = -k(500s)
ln(0.74) = -500k
k = (ln(0.74))/(-500s)
Calculating k using the given data, we find:
k ≈ -0.0011 s^-1
Now, let's use the rate law, the integrated rate law, and the value of the rate constant to predict the pressure of C2H5OH after 900s from the start of the reaction.
Using the integrated rate law:
ln(Pt/P0) = -kt
Rearranging the equation:
Pt = P0 * e^(-kt)
Substituting the values:
Pt = 250 torr * e^(-0.0011 s^-1 * 900s)
Pt ≈ 122.24 torr
Therefore, the pressure of C2H5OH after 900s from the start of the reaction is approximately 122.24 torr.
I can get you started.
rate for 100 s is 237/100 = 2.37
rate for 200 s is 224/100 = 2.24
Then rate1 = 2.37 = k(A)^x=k(237)
and rate2 = 2.24 = k(A)^x = k(224)^x
solve for x.
Then plug rate and (A) into any one of the trials and solve for k.