Arrhenius Graphs
1. Graphical Analysis
A reaction was completed at multiple temperatures and the rate constant was calculated for each. The information is in the table below. Using the data, draw a graph and use the graph to determine the activation energy of the reaction.
| Temperature / K | k / mol-1 dm3 |
|---|---|
| 200 | 3.15×10-7 |
| 250 | 4.30×10-4 |
| 300 | 5.22×10-2 |
| 350 | 1.65 |
| 400 | 19.7 |
| 450 | 161 |
Step 1: Process Data
Calculate 1/T and ln(k) values:
| T / K | 1/T / K-1 | k | ln(k) |
|---|---|---|---|
| 200 | 0.00500 | 3.15×10-7 | -14.97 |
| 250 | 0.00400 | 4.30×10-4 | -7.75 |
| 300 | 0.00333 | 5.22×10-2 | -2.95 |
| 350 | 0.00286 | 1.65 | 0.50 |
| 400 | 0.00250 | 19.7 | 2.98 |
| 450 | 0.00222 | 161 | 5.08 |
Step 2: Plot Graph
Plot ln(k) against 1/T. The graph should be a straight line with a negative gradient.
Step 3: Calculate Ea
Gradient ≈ -7204
Ea = -Gradient × R
Ea = -(-7204) × 8.31
Ea = 59.9 kJ mol-1
A reaction was completed at multiple temperatures and the rate constant was calculated for each. The information is in the table below. Using the data, draw a graph and use the graph to determine the activation energy of the reaction.
| Temperature / K | k / mol-1 dm3 |
|---|---|
| 298 | 9.61×10-27 |
| 338 | 9.02×10-25 |
| 378 | 3.24×10-23 |
| 418 | 5.84×10-22 |
| 458 | 6.37×10-21 |
| 498 | 4.73×10-20 |
Step 1: Process Data
Calculate 1/T and ln(k) values:
| T / K | 1/T / K-1 | k | ln(k) |
|---|---|---|---|
| 298 | 0.00336 | 9.61×10-27 | -59.9 |
| 338 | 0.00296 | 9.02×10-25 | -55.4 |
| 378 | 0.00265 | 3.24×10-23 | -51.8 |
| 418 | 0.00239 | 5.84×10-22 | -48.8 |
| 458 | 0.00218 | 6.37×10-21 | -46.5 |
| 498 | 0.00201 | 4.73×10-20 | -44.5 |
Step 2: Plot Graph
Plot ln(k) against 1/T.
Step 3: Calculate Ea
Gradient ≈ -11400
Ea = -Gradient × R
Ea = -(-11400) × 8.31
Ea = 95.1 kJ mol-1
A reaction was completed at multiple temperatures and the rate constant was calculated for each. The information is in the table below. Using the data, draw a graph and use the graph to determine the activation energy of the reaction.
| Temperature / K | k / mol-1 dm3 |
|---|---|
| 250 | 0.93 |
| 275 | 3.8 |
| 300 | 12.3 |
| 325 | 33.2 |
| 350 | 77.8 |
| 375 | 162.7 |
Step 1: Process Data
Calculate 1/T and ln(k) values:
| T / K | 1/T / K-1 | k | ln(k) |
|---|---|---|---|
| 250 | 0.00400 | 0.93 | -0.07 |
| 275 | 0.00364 | 3.8 | 1.34 |
| 300 | 0.00333 | 12.3 | 2.51 |
| 325 | 0.00308 | 33.2 | 3.50 |
| 350 | 0.00286 | 77.8 | 4.35 |
| 375 | 0.00267 | 162.7 | 5.09 |
Step 2: Plot Graph
Plot ln(k) against 1/T.
Step 3: Calculate Ea
Gradient ≈ -3920
Ea = -Gradient × R
Ea = -(-3920) × 8.31
Ea = 32.6 kJ mol-1