Tuesday

September 30, 2014

September 30, 2014

Posted by **Angie** on Tuesday, September 25, 2012 at 4:24pm.

My kelvin temperatures are:

298

308

318

328

338

348

358

368

378

Taking the natural log of both sides of both sides of the equation gives ln(k) = ln(A) – E/RT. Thus, a graph of the natural log of the reaction rate constant versus the inverse of the absolute temperature should be linear. Create such a graph to verify this is true.

- Chemistry -
**DrBob222**, Tuesday, September 25, 2012 at 4:27pm298 = T

inverse is 1/T = 1/298

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