The Arrhenius equation is k = Ae^(-Ea/RT)

Where k is the rate constant, E_a is the activationenergy, R is the ideal gas constant (8.314 J/mole*K) and T is theKelvin temperature.

The slope i got for my trend line is 33.3x+-4.74

and my R2 value is 0.922

Use the slope of the trendline to calculate the ActivationEnergy (Ea) of the reaction.

Activation Energy =

5. Using your results, calculate how long it would take for thereaction to occur at 0^oC.

Time for Reaction to Occur at0^oC =______________________________

These were the steps needed to be taken. The Y axisneeds to be Ln(1/t) as I made it above

Transfer the time/temperature data to an Excel spreadsheet.

2. Use the operations in Excel to calculate ln(1/?t) and 1/T(remember T must be in Kelvin).

3. Plot ln(1/?t) vs 1/T in a scatter plot format (do not connectthe data points) and apply a linear

Display the trendline, the equation for thetrendline and the R² value on the graph.

4. Use the slope of the trendline to calculate the ActivationEnergy of the reaction.

Activation Energy and slope. Can someone possibly help solve for this and show work I am having trouble. I calculated for my slopeas seen in the picture.

 

Transcribed Image Text:

Temperature (K) 284.25 294.15 304.15 312.65 321.15 Time (sec) 112 61 33 23 15 1/Temperature (1/K) 0.003518 0.0033996 0.0032879 0.0031985 0.0031138 Ln (1/Time) 1/Time (1/sec) 0.008928571429 -4.718498871 0.01639344262 -4.110873864 0.0303030303 -3.496507561 0.04347826087 -3.135494216 0.06666666667 -2.708050201 Ln (1/Time) vs. 1/Time (1/sec) Ln (1/Time) 0 -1 -2 -3 -4 -5 0.01 0.02 Ln (1/Time) 0.03 33.3*x + -4.74 R² = 0.922 0.04 1/Time (1/sec) (K) 0.05 0.06 Temperature (K) 284.25 294.15 304.15 312.65 321.15 Time (sec) 112 61 33 23 15 1/Temperature (1/K) 0.003518 0.0033996 0.0032879 0.0031985 0.0031138 Ln (1/Time) 1/Time (1/sec) 0.008928571429 -4.718498871 0.01639344262 -4.110873864 0.0303030303 -3.496507561 0.04347826087 -3.135494216 0.06666666667 -2.708050201 Ln (1/Time) vs. 1/Time (1/sec) Ln (1/Time) 0 -1 -2 -3 -4 -5 0.01 0.02 Ln (1/Time) 0.03 33.3*x + -4.74 R² = 0.922 0.04 1/Time (1/sec) (K) 0.05 0.06

Answer rating: 100% (QA)

To calculate the activation energy Ea of the reaction using the slope of the trendline you can use t... View the full answer