What fraction of the molecules in a gas at 300 K collide with an energy equal to or greater than Ea when Ea equals 50 kJ/mol? What is the value of this fraction when Ea is 100 kJ/mol?

Video transcript

Hello. Everyone in this problem being told to find the fraction of the molecules in the gas that will collide with the energy equal to or greater than the activation energy of 500 Kelvin's. And if the activation energy at the same temperature is 160 killer jewels per mole determine the value of this fraction. So for the fraction of molecules equation that is F. Equals to E raised to the power of negative E. Of a divided by R. Of T. So for our energy of activation, so E. Of A. That is equal to 80 killed jules Permal At the temperature equaling 2 500 Kelvin's. So we plug it into our fraction here. You see that we have erase the part of negative 80 killer jules Herrmann and this all to fight it by Let's see for our we have 8.314 times 10 to the negative three killer joules per mole times kelvin. And then for our T. Value eri said that is 500 kelvin's. So once you put that into the calculator received, that this fraction here is equal to 4.39 times 10 to the negative nine. So this is at the condition where the activation energy is at 500 kelvin's. So this is one part of our answer here. Now for the second part we have the same creation here. So F. Is equal to E raise a part of E. Of a over r of T. So four. When the energy of activation is equal to 160 killer jules Permal At the same temperature. So when T. is equal to 500 Kelvin's again, we'll go ahead and plug this into our equation here. So F is equal to erased your power of We have the e. of a equaling to 160 killed jules thermal. It's all divided by. So for our we have again, 8.314 times 10 to the negative three kila jules per mole times kelvin. And then for a T. Value, that's the exact same thing. So we have 500 kelvin's. So once you put that in the calculator, receive that this fraction is going to be equal to 1.92 times 10 to the negative 17. So this right here is going to be my final and last answer for this problem.