A reaction has a rate constant of 0.000122/s at 27 °C and 0.228/s at 77 °C. b. What is the value of the rate constant at 17 °C?

Question

Accepted Answer

hi everyone for this problem, it reads the rate constant values for a hypothetical reaction are 1.356 times 10 to the negative three seconds in verse and 2.412 times 10 to the negative three seconds inverse at 35 degrees Celsius and 43 degrees Celsius, respectively, calculate the value of the rate constant at 52 degrees Celsius. So our goal here is to calculate the value of the rate constant at 52 degrees Celsius. So let's take a look at our problem to see what were given were given to rate constants here. So this is the first one and we can call this K one and this one is our second rate constant, which we can call K two. We're also given to temperatures that go along with those rate constants. So we'll say this is T one and we'll say this one is T to Ok, so the first thing we're going to want to do is convert these two kelvin because they're in degrees Celsius. And to convert this to kelvin, we're going to add 273.15. So that gives us a new T one Of 308.15 Kelvin and a new T two of 316. Kelvin. Okay, so because we're given to rate constants at two different temperatures, we can solve for activation energy and once we get that activation energy we can calculate the value of the rate constant at 52 degrees Celsius. So let's go ahead and do that first to solve for activation energy When we're given to rate constants at two different temperatures, we can use this equation. So that is Ln of K two over K one is equal to negative activation energy over our which is our universal gas constant times one over T two minus one over T one. So by using this equation will be able to solve for activation energy, which is our goal here. So let's go ahead and plug in what we know based off of what was given in the problem. So for L N F K two over K one, we said K two is 2.412 times 10 to the negative three seconds inverse, and K one is 1.356 times 10 to the negative three seconds inverse. Okay, so that's the left side. Our right side is going to be negative activation energy, which is what we're solving for. R Universal gas constant. R is 8. jewels over mole times kelvin. This is a value. We should know. It wasn't given in the problem. Now for our temperatures we have one over T two Which is 316. Kelvin -1 over T1, which is 0.15 kelvin. Okay, so now you can see why we converted our degrees Celsius to kelvin. It's because our universal gas constant R has the units of kelvin. So if we use Celsius, it wasn't going to work. So let's go ahead and simplify this a little bit. Our left side becomes 0.5759 is equal to our right side is activation energy times 9.877 times to the negative six. Okay, so because we want to solve for activation energy, we're going to divide both sides by 9.877 times 10 to the negative six. So we get activation energy is equal to 0.5759, divided by 9.87, 7 times 10 to the -6. So we get an activation energy of 58, 307 Jules Permal okay, so now that we know what our activation energy is, we can use this activation energy to calculate the value of K at 52°C. So we're going to use the same equation that we just had above but our values are going to change a little bit, so I'm going to copy this equation here. Okay, so we get Instead of solving for activation energy now we're going to be solving for one of our case which is going to be K two. Okay, so we're right out what we know. So we know that K one is equal to two point, we can choose either of the first two K values we had in our equation or our reaction above to use for K one. So we'll use 2.412 times 10 to the negative three seconds inverse. R T one which goes with K one was 316.15 kelvin. So now our K two is what we want to solve for. Because it's asking us to calculate the rate constant at 52 degrees Celsius. So that means T two is equal to 52 degrees Celsius. But remember we need to convert this to kelvin. So we're going to add 273.15 which gives us AT two of 325. Kelvin. Okay. And we already know what our activation energy is and we know what our our is. So our activation energy we just solved for is 58,307 jewels per mole. Okay. And our our Is 8.314 jewels Permal. Okay, so we have everything we need to plug in and solve for K two. So let's go ahead and do that. So we get L. N. Let me move this over to the side here. Okay, so this is everything that we know. So that means L. N. Of K two over K one is going to be K two over K one is 412 times 10 to the negative three seconds inverse Is equal to negative activation energy. So we have negative 58, 307 jules per mole. And this is over R universal gas constant 8.314 jewels over mole times kelvin. And this is going to be multiplied by one over T two minus one over T one. So 325. kelvin minus one over T one. Which is 316. Kelvin. Okay, so let's go ahead and simplify this here. The left side stays the same because we're solving for K two. So we'll leave that the same for right now and the right side becomes negative 7013 times negative 8.77 or seven. Excuse me? 55 Times 10 to the negative five. Okay, so let's simplify the right side some more. So we get 0. 40 is going to equal RLN of K two over R K one. All right, so there we go. Alright, so in order for us to get rid of the L. N, we're going to need to raise both sides to E. Okay, so when we raise both sides to E, this Ln is going to cancel here. So we're just going to get K two over 2.412 times 10 to the negative three seconds in verse and E raised to the 30.6140 is going to be 1.848. Okay, so now we can multiply both sides of our equation by 2.412 times 10 to the negative three. And when we do that we get K two is equal to 4.457 times 10 to the negative three seconds inverse. And this is going to be our final answer. This is the value of the rate constant at 52 degrees Celsius. That's the end of this problem. I hope this was helpful.

A reaction has a rate constant of 0.000122/s at 27 °C and 0.228/s at 77 °C. b. What is the value of the rate constant at 17 °C?

Verified Solution

Key Concepts

Rate Constant

Arrhenius Equation

Temperature Conversion