 Do subscribe to Ekeeda Channel and press bell icon to get updates about latest engineering HSC and IIT JEE main and advanced videos hello Friends in the last lecture we have also did the first-order reaction as the rate law for the first order reaction but now we are going to do the rate law for zero order reaction so let us see how can we derive it as we know that the reactant gives us pearl and as the eye increases the consideration of reactant goes on decreases so for that thing I would consider the reactant a which is giving me the product so I’m considering a reaction in which a is giving me product so as the type increases the concentration of reactant goes on decreasing so this is nothing but the rate of fashion so for this I could write the rate of H and s 3 which is equals to minus concentration of da divided by DT the – symbolizes that the concentration decreases with respect to a and now since we are talking about it’s zero order reaction so we could write the zero order reaction in the form of rate law also so let us see ready to rate low rate is nothing but K into concentration of in that is reactive but since it is a zero order reaction so the experimental value that is the value of X component in this case will be zero so let us see what as we can do with this thing so now the equation can be written as 3 is equals to K but we have also got the rate array also which is this one so I would mark this equation as equation 1 and the equation that we have got right now as equation 2 so therefore equating both the equations 1 and 2 will get so equating equation 1 and 2 we get minus da upon DT is equals to K so I can do the further process like minus da is equals to K 2 DT and I’ll show the negative sign to one side that is d is equals to minus K of DT but this equation is a differential equation for a zero order reaction but we have to integrate it so let us see what happens when we are going to integrate it so indicating the equation we could get this thing but integration should be in a limit that is a lower limit and a higher limit so I would consider whether time these goes to 0 the concentration of the reactant a would have been a naught and then when time T goes to P then the concentration of reactant would have been 80 so since we know that integration of DX is nothing but X so similarly the integration of da will be a but with a limits that is with a upper limit then a is 80 and when a lower limit when a is you know similarly we can apply a limit for the time also so we could get when time T goes to e and when lower limit when T goes to C so after doing this equation what we get is instead of a we will write the upper limit as 80 minus the lower limit that is concentration when it is 80 minus the concentration of phaeton when it is you know is equals to okay again when upper limit is e and the lower limit is 0 so we get finally yes a P minus a 0 is equal to minus K of e so what we are going to do is now we are going to relate with the rate constant and the further part that is the further parameters like is U and 80 so let us do it so by arranging the above equation that we have got we could write the equation as KT is equals to minus of 80 plus of a dot that could be written as a dot minus 80 simple and this equation can be finalized later s KS goes to a naught minus 80 divided by T so this is the final equation that we have got or we have made a relation between the rate constant for a zero order reaction I hope you have liked this video and you have got to know that how to derive an equation for the rate constant for a zero order reaction so thank you friends for watching this video I hope you have liked this video and you will share with your friends too and please don’t forget to subscribe Nikita children thank you so much

• ### Ankush Chauhan

Nice sir

• ### Rs Singh

Tqq so muchhh sir

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