The activation energy of an uncatalyzed reaction is 95 kJ/mol. The addition of a catalyst lowers the activation energy to 55 kJ/mol. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) 25 C

Video transcript

Hello everyone today. With the following problem, the activation energy of an un catalyzed reaction is 95 kilojoules per mole. The addition of a catalyst lowers the activation energy to 55 kilojoules per mole. Assuming that the collision factor remains the same. By what factor will the catalyst increase the rate of the reaction at 25 °C. So we can find the rate constants for both reactions at 25 °C. And they compare their ratios to find a factor by which the catalyzed reaction is faster. But first, we have to convert our temperature from degrees Celsius to degrees Kelvin. And to do that, we simply just add 2 73.15 to give us 2 98.15 degrees Kelvin. Now we will use the AUS equation to calculate the rate constants for both reactions. So for our first reaction, we have the following and for our second reaction, it will be very similar except we will be using an activation energy specific to that second reaction. No. As in the question, the collision factor remains the same for both reactions at the same temperature. So we can actually just find the ratio with their constants. So we can simplify these two into the following. So we have the reaction of K of the second reaction over the first, that of the first reaction is equal to E raised to the power of the activation energy for reaction two subtracted by that of reaction one. And then we divide that by the R which is a gas constant, we multiply that by temperature. So let's not plug in the values and solve. So we have E as raised to the negative value. So we have our first reaction activation energy of reaction to which was 55 kilojoules per mole. And what we want to do is once we, once we actually subtract that from the activists energy of our first reaction, we can convert these units from kilojoules to jewels using a conversion factor that 10 raised to the power of three is equal to one kilojoule or 10 to the power of three joules is equal to one kilojoule. We then divide that by gas constant, which can be found in a reference text which is 8.314. You multiply that by the temperature in Kelvin 2 98.15 degrees. Simplifying this even further, we get E is raise to the power of 40,000, divided by 2475.5 plugging this into our calculator and solving further to simplify this ratio, we get E raise of the power of 16.143 which is also equal to 10,410,774. Now, of course, looking at our anchos, we see that and choice B is best reflective of this. The catalyze reaction is 10 million times faster at 25 °C. And with that, we have solved the problem overall, I hope this helped. And until next time.