The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. If an enzyme increases the rate of the hydrolysis reaction by a factor of 1 million, how much lower must the activation barrier be when sucrose is in the active site of the enzyme? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical and a temperature of 25 °C.)

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Hi everyone for this problem we're told the activation energy for the hydraulic sis of your area under atmospheric conditions is 60.93 kg per mole. However, the uris enzyme boost the reaction rate by a factor of 13.6 Six million, calculate the difference in activation energy between the catalyzed and catalyzed reactions. Assume that the frequency factor for both reactions is the same at 25°C. Okay, so our goal here is to calculate the difference and activation energy. Okay. And we can do this by using our Iranians equation. And that equation is K. Is equal to our frequency factor A times E raised to our negative activation energy over R. T. So we're given to two K values here. So we'll say that K1 is going to represent our uncapped realized reaction and K two is going to represent are catalyzed. Okay. And we know what the rate is between these two because they tell us that it is 13.6 million. So that's the reaction rate. So our K two over K one is 13. million. And if we write that out in exponent form it's 13.6 times 10 to the sixth. Okay, so our goal here is to use this Iranians equation to solve for our difference in activation energy. So what we can do is we can say K two over K one is equal to are active. Are frequency factor times raised to the E. For our negative activation energy over R. T. For both of them. Okay, so we'll just go ahead and copy that down. Okay, in the problem they tell us to assume that the frequency factor for both reactions is the same. So A is our frequency factor. So because they're the same, we can go ahead and cross this out. So what this simplifies to is K. Two over K. One equals E. To the negative activation energy over R. T. So to get rid of R. E. Exponents, we're going to raise both sides of this equation by Ellen. So let me write that in a different color. So Ln of K two over A. Of K one is equal to Ln of the right side. When we do that our E. Is going to cancel our exponent is going to cancel. And so what this simplifies to is Ln of K two over A. K. One is equal to negative activation energy over R. T. So we can say this is activation energy to over Negative activation energy one over R. T. So let's go ahead and simplify this. We can bring this denominator up so we can get rid of this fraction on the right side. So when we bring the denominator up we get Ln of K two over K. One is equal to negative activation energy to over R. T. Plus activation energy one over R. T. So we simplified that by making by bringing the denominator up to make it positive. So let's go ahead and factor out R. T from both of these. Okay, so when we factor out our T. We get Ellen of K two over K one is equal to activation energy one minus activation energy to over R. T. Okay, so now that we factored R. T. Let's multiply both sides of our equation by R. T. So that we can just have activation energy one minus activation energy too. So let's multiply both sides by R. T. And when we do the R. T. On this side cancels and we'll be left with activation energy one minus activation energy two is equal to R. T. Times Ln of K two over K one. And this is going to give us the energy. I mean the answer for the difference in activation energy. Okay, so let's go ahead and plug in because we have everything we need. So when we plug everything in we'll get activation energy one minus activation energy two is equal to R. Is r universal gas constant. This is a value that's not going to be given but we should know it's 8. jewels over mole times kelvin. This is going to be times our temperature. So our temperature was given To be 25°C. So we need to convert this 25°C to Kelvin and to do that we're going to add 273.15. So when we do that we get a temperature in Kelvin of 298.15 Kelvin. So let's go ahead and plug that in for our temperature. So we have to 98 15 kelvin and this is going to be times the Ln of K two over K one, which we were told it's 13.6 million and we converted that to exponents. So this is going to be Ln of 13.6 times 10 to the sixth. So once we calculate this, we're going to get a difference of 40, 716 jules per mole. So let's go ahead and convert this to kill a jewel so we can have a nicer number. So this 40, jules per mole And one killer jewel There is 1000 jules. So our jewels cancel and we'll be left with killer jewels Permal. So we get 40. killer jewels per mole. And this is going to be our final answer. This is going to be the difference in activation energy between the catalyzed and catalyzed reactions. That's the end of this problem. I hope this was helpful

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Key Concepts

Activation Energy

Enzyme Catalysis

Arrhenius Equation