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The rate constant of a first-order reaction is $4.60 \times 10^{-4} \mathrm{s}^{-1}$ at $350^{\circ} \mathrm{C} .$ If the activation energy is $104 \mathrm{kJ} / \mathrm{mol},$ calculate the temperature at which its rate constant is $8.80 \times 10^{-4} \mathrm{s}^{-1}$.

$$371^{\circ} \mathrm{C}$$

Chemistry 102

Chapter 6

Chemical Kinetics

Kinetics

University of Central Florida

Brown University

University of Toronto

Lectures

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In probability theory, the conditional probability of an event A given that another event B has occurred is defined as the probability of A given B, written as P(A|B). It is a function of the probability of B, the probability of A given B, and the probability of B.

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In chemistry, kinetics is the study of the rates of chemical reactions. The rate of a reaction is the change in concentration of a reactant over time. The rate of reaction is dependent on the concentration of the reactants, temperature, and the activation energy of the reaction.

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the rate constant of a fir…

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A certain first-order rea…

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well, so they were going to be looking at a situation where were given the great constant at 350 degrees Celsius and the activation energy. We want to find the temperature where the rate constant will be. This value right here. So how can we do that? While remember, if we graph the night Chong of K over one over time temperature, this slope will be equal to negative activation energy over our So if we know the slope and were given one point when we confined the other point to So this right here, that will be this point right here. So now let's try to create an expression for the soap. So we have the natural log of 4.60 times 10 to the negative four minus the natural log of this point right here. This Kate, this K is smaller than this case, So the larger K is would be this point right here. So we're going to write it like this. And now we're going to write the one over tea Soto rendered in a 50 degrees Celsius in Kelvin. We would just at 273 so that would be 623 and then our slope is given by the negative of the activation energy over the are which is 8.3 14 So now we just have to solve this. So it's solved the top part there and we will see that it will be negative 0.6 49 No and then over here it'll be and we can just erase the negative signs on both sides. We will see that this side will be in this. So now let's simplify it further. So you're going tohave 20 subtracting 1005 109 over the second temperature. So now keep going. And if we keep solving, we will see that our second temperature will be 644 Killman and if we want to convert that into degree Celsius, we would just subtract 223. We will see that would be 300 in 71 degree Celsius.

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