ahoy mateys let's find the integrated rate law for a first-order reaction a first-order reaction has the rates proportional to both a rate constant K and the concentration of a linearly rather the exponent is 1 that was that's what makes it a first-order reaction to get the integrated rate law we are going to need to rewrite rate as the rate of disappearance of a over time this here is the definition of rate and again it's proportional to the rate constant times the concentration of a separate your variables is the next step what that means is I'm going to leave da on the left and move the concentration of a tiller to the left as well notice it was in top now it's in the denominator because I had to divide both sides by a and on the other side I'm going to move my negative I'm going to leave my K there now I'm going to move my DT to the right as well notice it was in the denominator now it's in the numerator because we had to multiply both sides by DT in order to get it over there now that I'm here I'm actually literally going to integrate both sides get it integrated rate law let me show you more accurately what I'm going to end up doing I'm going to integrate 1 over a DA on the left and I'm going to integrate 1 DT on the left it's a rule of integrals that you're allowed to move constants outside of it I'm left with one here the same way if you factor an X out of X you're left with one anyways what's the integral of 1 over a I asked you you tell me the answer is lon of a I asked you what the integral of one is and you tell me it is T hey look you're right because you're smart it goes from zero to T in both cases if you're familiar with integration you're familiar with this notation that I'm using what that really means is that the lawn of a at t minus the lawn of a at zero is equal to negative K T at time t minus T at time zero obviously T is zero at time zero beautiful now what I'm going to do is rearrange this for you just so you can emphasize what exactly is going on lon a of T is negative KT multiplying through plus the lawn of a knot what this means is that if i graph lawn of a on my y-axis and T on my x-axis I'm going to end up with a straight line with a slope of negative K and a y-intercept of lon a knot let me emphasize what I mean I put time on my x and i graph lawn of concentration on my Y I'm going to get a straight line the slope is going to be negative K and the y-intercept will be at the lawn of a knot how awesome is that the whole reason we do integrated rate laws is to turn something regular and boring like this into something awesome like this I want to show you one more thing that only pops out in the first order integrated rate law and that is I could rearrange this another way to get a of T equals this markers crap a knot e to the minus KT that's using logarithm rules over here undoing lawn with E ^ and then multiplying by eat both sides by a knot all I want to emphasize here is that a varies exponentially or rather negative exponentially or whatever you want to call it with T a decays exponentially over time that's the wording I was looking for beautiful you guys should try doing this on your own make sure you understand each step especially if your teacher actually cares about you doing it and until then best of luck to you
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