Given that the initial rate constant is 00180s1 at an...

Given that the initial rate constant is 0.0180 s^-1 at an initial temperature of 27 °C, what would the rate constant be at a temperature of 200 °C for the same reaction described in Part A? Part A: The activation energy of a certain reaction is 42.1 kJ/mol. At 27 °C, the rate constant is 0.0180 s^-1. At what temperature in degrees Celsius would this reaction go twice as fast?

Transcript

00:01 Here we need to find out the iopic name of the given compound.

00:04 So here what we have is c, here we have c, triple one.

00:09 Here we have c, then we have c and c.

00:12 Here we have the c.

00:13 Here first what we need to do is we need to put the required number of the hydrogen atom to each carbon.

00:20 So as we know the carbon needs the four atoms attached to it, whether it is hydrogen or any other atom.

00:28 Here we need to put the hydrogen atoms so here this carbon is having the one bond here this carbon is having one bond so here what we will get is we will get the three hydrogens attached to it now if we talk about this carbons where we have one two three four bonds done so here no carbon now here what we will have here again four bonds no hydrogen will be there now if we talk about these two these carbons so here this carbon has three bond so we will get one hydrogen here this carbon has one bond so again here we will have the three hydrogen here this carbon one one one two three hydrogen so we have placed the hydrogen so it will be like this h 3 c here we have c triple bond c c h here we have c h3 and here also we have c h3 so this is the required alkyne here what we have the triple bond so this is alky now we need to give the iupac name so here we have the triple bond so it will be given the preference before that we will see the longest chain if we start from here 1 2 3 so if we are starting from here this carbon the triple bond carbon is getting number 3 but we need to give it to the least number so if we will start from here 1 2 3 4 5 so from here 1 2 3 4 5 again here we are getting number 3 so we will start the numbering from this carbon.

02:01 1, 2, 3, 4 and 5 because the longest chain is forming of the 5 carbon only.

02:09 But now we will see that which numbering is giving the least number to the main bond.

02:17 That is a triple bond.

02:18 So that is number 2 and this numbering...

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