Variation of the rate constant with temperature for the first-order reaction

2N2O5(g)  2N2O5(g) + O2(g)
Is given in the following table. Determine graphically the activation energy for the reaction.

T(K) k(s^-1)
2981.74 x 10^-5
308 6.61 x 10^-5
318 2.51 x 10^-4
328 7.59 x 10^-4
338 2.40 x 10^-3

103.0 kJ/mole

Plotting lnk vs 1/T produces a line. The slope will be Ea and the intercept will be A. You can read more about it here.

http://en.wikipedia.org/wiki/Arrhenius_equation

Scroll down about half way on the page.

Well, well, well, looks like we're dealing with some chemistry here! Let's dive into it.

To determine the activation energy graphically, we need to plot the rate constant (k) against the reciprocal of the temperature (1/T). The slope of this plot will give us the activation energy.

So, let's draw the plot. But remember, I can't draw it for you, so just imagine a lovely graph in front of you.

On the x-axis, we have the reciprocal of the temperature (1/T), and on the y-axis, we have the rate constant (k).

Plotting the given values, we connect the points with a smooth line.

Now, here comes the interesting part - drawing a line of best fit. We need to find a straight line that fits the trend of the plotted points. Make sure to avoid any crooked lines, we want a handsome, straight line!

The trick is to choose two points on the line for the steepest part of the curve (the warmest part). Then, we can draw a straight line that intersects these two points. Let's call this line "Line of Best Fit."

Now, let's calculate the slope of this line. Slope in this case is nothing but the negative of the activation energy divided by the gas constant (R). It's like sliding down a slope, but with fancy math involved.

Once we have the slope, we can easily find the activation energy by multiplying it with the gas constant (R).

Now, the value of the gas constant (R) is approximately 8.314 J/(mol⋅K). Keep this magical number in mind.

So, take the slope we just calculated from the graph, multiply it by the gas constant, and voila! You've got the activation energy staring right at you.

Hope this explanation didn't make you "react" too much! Good luck with your chemistry adventures!

To determine the activation energy for the reaction, we can use the Arrhenius equation:

k = A * e^(-Ea / (RT))

Where:
k is the rate constant,
A is the pre-exponential factor,
Ea is the activation energy,
R is the ideal gas constant (8.314 J/(mol*K)),
and T is the temperature in Kelvin.

We can take the natural logarithm of both sides of the equation:

ln(k) = ln(A) - (Ea / (R * T))

Now, let's calculate the natural logarithm of the rate constants at each temperature given in the table:

T(K) k(s^-1) ln(k)
298 1.74 x 10^-5 -11.945
308 6.61 x 10^-5 -9.710
318 2.51 x 10^-4 -8.282
328 7.59 x 10^-4 -7.178
338 2.40 x 10^-3 -6.035

Next, we'll graph ln(k) against 1/T:

T(K) 1/T(K) ln(k)
298 0.00336 -11.945
308 0.00325 -9.710
318 0.00314 -8.282
328 0.00305 -7.178
338 0.00296 -6.035

Now, we can plot the points on a graph with ln(k) on the y-axis and 1/T on the x-axis. Make sure the temperature values are converted to Kelvin.

By fitting a straight line to the data points, we can determine the slope of the line, which is equal to -Ea / R. Rearranging the equation, we can solve for the activation energy Ea:

Ea = -slope * R

Where slope is the slope of the line obtained from the graph, and R is the ideal gas constant (8.314 J/(mol*K)).

So, by determining the slope of the line, we can calculate the activation energy for the reaction with the given data.

To determine the activation energy for the reaction graphically, we can use the Arrhenius equation. The Arrhenius equation relates the rate constant (k) with temperature (T) and the activation energy (Ea) of the reaction:

k = A * e^(-Ea/RT)

where:
- k is the rate constant
- A is the pre-exponential factor (also called the frequency factor)
- Ea is the activation energy
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin

To find the activation energy graphically, we need to plot ln(k) against 1/T and determine the slope of the line. The slope of the line is equal to -Ea/R.

Let's calculate ln(k) and 1/T for each temperature in the table:

T(K) k(s^-1) ln(k) 1/T (K^-1)
298 1.74E-5 -11.90 0.00334
308 6.61E-5 -9.71 0.00325
318 2.51E-4 -8.35 0.00314
328 7.59E-4 -7.18 0.00305
338 2.40E-3 -6.03 0.00296

Next, plot ln(k) against 1/T on a graph. The x-axis will be 1/T, and the y-axis will be ln(k). Fit a line through the data points and determine the slope. The slope of the line will give us the activation energy.

Once you have plotted the graph, calculate the slope using the equation:

Slope = (ln(k2) - ln(k1)) / (1/T2 - 1/T1)

Choose any two points on the line, for example, (0.00334, -11.90) and (0.00296, -6.03).

Slope = (-6.03 - (-11.90)) / (0.00296 - 0.00334) = 5.87

Now that we have the slope, we can calculate the activation energy using the slope and the gas constant (R = 8.314 J/mol·K):

Activation energy (Ea) = -slope * R

Ea = -5.87 * 8.314 = -48.75 kJ/mol

Remember to use the appropriate units for the gas constant (R) and activation energy (Ea) based on the units of k and temperature in the Arrhenius equation.

Therefore, the activation energy for the reaction is approximately 48.75 kJ/mol.