Saved Bookmarks
| 1. |
Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation log k = log A \(\frac{E_a}{2.303R}\) \(\Big(\frac{1}{T}\Big)\)Where Ea is the activation energy. When a graph is plotted for logk Vs \(\frac{1}{T}\) a straight line with a slope of- 4000K Is obtained. Calculate the activation energy. |
|
Answer» log k = log A - \(\frac{E_a}{2.303R}\) \(\Big(\frac{1}{T}\Big)\) y = c + mx m = -\(\frac{E_a}{2.303R}\) Ea = -2.303Rm Ea = – 2.303 x 8.314 J K-1 mol-1 x (-4000 K) Ea = 76,589 J mol-1 Ea =76589kJmol-1 |
|