The half life of a certain reaction with second order was found to be, oh, 0.45 seconds. What was the initial concentration of a reactant if the slope of the straight line for this reaction is 3.5 * 10 to the -3? All right, so they tell us it's second order, which means that half life equals 1K×initial concentration.

Here we need to solve for initial concentration. So what we do is we multiply both sides by K times initial concentration, which becomes K times initial concentration times half life equals one. Divide both sides now by K and half life and we've isolated our initial concentration. So initial concentration equals 1rate constant K times half life.

All right. So now, knowing this, let's solve. So initial concentration equals 1rate constant K times half life. We're told that our half life is 0.45 seconds. But how do we figure out K? Remember when we dealt with the integrated weight loss? When it comes to K, it's equal to the slope of a straight line. So them telling us the slope of the straight line is them really telling us what K is.

So K here is this value. So we plug in 3.5 * 10 to the -3. Doing that gives us our initial concentration, which is 634.92 molar. Since both of our numbers have two sig figs, this comes out to 630 molar, so this would be our initial concentration of our reactant.