in this question, we have a reaction Where Direction Constant is 0.1171 over seconds at 400 Kelvin, and the reaction constant is 0.6891 over seconds at 450. Calvin. They're asking us in the first part of this question what is the energy barrier? Basically the energy off actuacion. And at the second part of this question, they're asking us the constant reaction Constant at 420 fire Calvin. So let's first go for energy of activation. What we're going to use is this formula. Ellen que two or K one equals two energy off activation or 8.314 one over T one minus one over teachings. Now, as we proceed, what we're going to write for K two is 0.689 over 0.117 This is equal to E. A, or 8.314 in Prentice's one over t. One is 400 minus. T two is 450 Now. If you sold this equation for E A, we get e A to B 12 19 85 Joel Sperm ALS. Now we need to solve the second part of this question, and we're going to use the same exact formula. Just in this case, R K is going to be for 425 instead off 450 over here. So Ellen say over 0.117 equals two R e a one 21985 divide by 8.314 one over 400 minus one over 400 25. Now, if you sold the right hand side off the equation and the left hand side off the equation, what we're going to get is Ellen Que over 0.117 equals two 2.157 Remember, if you take the into the power for both sides is going to be It's the power of two point 157 and from the math, eat the power off Ln x is going to be equal to X. So que over 0.117 equals two eight point 65 10 and this is going to be eat the part of 2.157 And if you multiply both sides by 0.117 You find Kay to be zero point 101 now, just for a little confirmation. Is this value a good value? So at 400 r k one is 0.0 once at 4. 50 r k. Value 0.6 almost seven. So at 4. 25 we should have a result that is, in between those values and our values in between those values. So that can pretty much confirm that we have the right value. And if you basically see a question like this where they're asking for the K for a different, uh, temperature, what you're supposed to do is first find everything that you can find. E A. If the question was somewhat different, maybe you should have found the a value and then proceed to the second, um, constant finding the second constant

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## Video Transcript

in this question, we have a reaction Where Direction Constant is 0.1171 over seconds at 400 Kelvin, and the reaction constant is 0.6891 over seconds at 450. Calvin. They're asking us in the first part of this question what is the energy barrier? Basically the energy off actuacion. And at the second part of this question, they're asking us the constant reaction Constant at 420 fire Calvin. So let's first go for energy of activation. What we're going to use is this formula. Ellen que two or K one equals two energy off activation or 8.314 one over T one minus one over teachings. Now, as we proceed, what we're going to write for K two is 0.689 over 0.117 This is equal to E. A, or 8.314 in Prentice's one over t. One is 400 minus. T two is 450 Now. If you sold this equation for E A, we get e A to B 12 19 85 Joel Sperm ALS. Now we need to solve the second part of this question, and we're going to use the same exact formula. Just in this case, R K is going to be for 425 instead off 450 over here. So Ellen say over 0.117 equals two R e a one 21985 divide by 8.314 one over 400 minus one over 400 25. Now, if you sold the right hand side off the equation and the left hand side off the equation, what we're going to get is Ellen Que over 0.117 equals two 2.157 Remember, if you take the into the power for both sides is going to be It's the power of two point 157 and from the math, eat the power off Ln x is going to be equal to X. So que over 0.117 equals two eight point 65 10 and this is going to be eat the part of 2.157 And if you multiply both sides by 0.117 You find Kay to be zero point 101 now, just for a little confirmation. Is this value a good value? So at 400 r k one is 0.0 once at 4. 50 r k. Value 0.6 almost seven. So at 4. 25 we should have a result that is, in between those values and our values in between those values. So that can pretty much confirm that we have the right value. And if you basically see a question like this where they're asking for the K for a different, uh, temperature, what you're supposed to do is first find everything that you can find. E A. If the question was somewhat different, maybe you should have found the a value and then proceed to the second, um, constant finding the second constant

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