The balanced equation for a given reaction is ( 4 X+3 Y...

The balanced equation for a given reaction is ( 4 X+3 Y ightarrow P ). The following data regarding initial reaction rates were collected in the lab. What is the rate law for the reaction? egin{tabular}{|l|l|l|l|} hline Trial & ( [X](M) ) & ( [Y](M) ) & Rate ( (M / s) ) \ hline 1 & 0.020 & 0.060 & ( 3.40 imes 10^{-3} ) \ hline 2 & 0.020 & 0.120 & ( 2.72 imes 10^{-2} ) \ hline 3 & 0.080 & 0.060 & ( 1.36 imes 10^{-2} ) \ hline end{tabular} Rate ( =k[X]^{4}[Y] ) Rate ( =k[X][Y] ) Rate ( =k[X][Y]^{3} ) Rate ( =k[X][Y]^{2} ) Rate ( =k[X]^{4}[Y]^{2} ) Rate ( =k[X]^{2}[Y]^{3} )

Transcript

00:01 Okay, so we have given here a chemical reaction.

00:03 Let me first write that this is 4x plus 3y gives p and now rate.

00:11 So by rate law, okay, we are writing the rate by rate law or rate equation.

00:20 So rate is equals to k multiplied by concentration of x multiplied by concentration of y so the concentration of reactants are to be considered here okay so now for trial one so now we have rate is equals to k multiplied by zero 0 .020 multiplied by 0 .060 okay now here it is to be keep in mind that the powers are also here so these power raised to power m and raised to power n so these are actually the order with respect to each concentration okay these orders are determined experimentally okay so now that's why here we can write raise to power m and raise to power n respectively also this rate value so that is also given to us so let me put that so this is 3 .40 multiplying 10 raised to power minus 3 okay so now consider this as first equation similarly now for trial 2 so now the rate is 2 .72 2 multiplying 10 raised to power minus 2 okay and that is equals to k multiplied by concentration of the reactant 0 .020 raised to power m 0 .120 raised to power n okay consider this as second equation okay now by 1 divided by 2 so this will be 3 .40 multiplying 10 raised to power minus 3 divided by 2 .72 multiplying 10 and raised to power minus 2.

03:35 And now k to k will be cancelled out 0 .020 molar sorry raised to power m.

03:44 And this will also be cancelled out.

03:46 Okay.

03:46 So let me write it again.

03:49 So this will be like 0 .020 divided by 0 .020 raised to power m.

03:56 We can consider the power as whole.

03:59 Right.

04:00 And this will be 0 .060 divided by 0 .120 raised to power n okay now solving this so this will be 1 .25 multiplying 10 raised to power minus 1 is equals to now solving this this will be 1 and now solving this this will be this is 0 .5 raised to power n okay now i can write this 1 .25 as shifting the decimal this can also be written as simply this is we have 0 .125 okay now this can also be further can also be written as 0 .5 raised to power 3 okay so if 0 .5 is multiplied three times we will get 0 .125 right and that is equals to 0 .5 raised to power n now when the base value is same so now the exponent values are equal to each other so now this can be written as or n is equals to 3 ok so that is the value for n now for trial 3 so we have information that 1 .36 6 multiplying 10 raised to power minus 2 that is equals to 0 .8 a multiplied by 0 .0 8 0 raised to power m multiplied by 0 .0 6 0 raised to power n okay consider this has third equation okay so this is the third equation let me write it again now by 3 divided by 1 now we can write that 0 point or let me write it again so this this is 1 .36 multiplying 10 raised to power minus 2 divided by 3 .40 multiplying 10 raised to power minus 3...

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