7E.3 in 7th edition
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7E.3 in 7th edition
the problem says "the presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kj to 75 kj, by what factor does the rate increase at 298, all other factors equal?" I tried to do this using the Kr2/Kr1= e^-(Ea2-Ea1)/RT but keep ending up with the wrong answer than the solutions manual. The solutions manual for some reason does -(.4*125)/RT for the exponent instead of -(125-75kj)/RT and i cant understand why, thanks for any help!
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Re: 7E.3 in 7th edition
125-175=50
0.4(125)=50
The two ways of writing it are equivalent. You must be making a mistake elsewhere.
0.4(125)=50
The two ways of writing it are equivalent. You must be making a mistake elsewhere.
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Re: 7E.3 in 7th edition
aisteles1G wrote:the problem says "the presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kj to 75 kj, by what factor does the rate increase at 298, all other factors equal?" I tried to do this using the Kr2/Kr1= e^-(Ea2-Ea1)/RT but keep ending up with the wrong answer than the solutions manual. The solutions manual for some reason does -(.4*125)/RT for the exponent instead of -(125-75kj)/RT and i cant understand why, thanks for any help!
Hi! Is this equation given to us on the sheet?
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- Posts: 117
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Re: 7E.3 in 7th edition
Ruiting Jia 4D wrote:aisteles1G wrote:the problem says "the presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kj to 75 kj, by what factor does the rate increase at 298, all other factors equal?" I tried to do this using the Kr2/Kr1= e^-(Ea2-Ea1)/RT but keep ending up with the wrong answer than the solutions manual. The solutions manual for some reason does -(.4*125)/RT for the exponent instead of -(125-75kj)/RT and i cant understand why, thanks for any help!
Hi! Is this equation given to us on the sheet?
Not exactly, this equation from the problem is just using the K=Ae^(-Ea/RT) equation for each K, which is on the equation sheet and you have to divide the Kr2/Kr1 and your A cancels out and you get the exponent with the two activation energies, so just remember to put the K2 one on top when dividing!
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