The activation energy for a certain reaction is 113 KJ/mol. By what factor (how many times) will the rate constant increase when the temperature is raised form 310K to 325K?
Answer: 7.6
Well, let me explain it in a way that won't make you feel like you need an activation energy just to understand the answer. So, when you increase the temperature from 310K to 325K, it's like giving the reaction a nice warm hug. And just like humans, reactions love warm hugs!
Now, the rate constant is like the speed at which the reaction happens. So, when you give the reaction a warm hug by increasing the temperature, it gets all excited and starts moving faster. It's like a high-speed chase, but without the cops and the action movie soundtrack.
To calculate how much faster the reaction will be, we can use the Arrhenius equation. Now, PLEASE don't run away screaming just because I mentioned an equation. It's really not that scary, I promise.
The Arrhenius equation tells us that the rate constant (k) is proportional to the exponential of the negative activation energy (Ea) divided by the product of the gas constant (R) and the temperature (T). Math, math, math, I know, but bear with me.
So, if we plug in the activation energy at 310K and 325K into the equation, we can compare the rate constants. And the difference turns out to be about 7.6 times!
So there you have it, my friend! By increasing the temperature from 310K to 325K, the rate constant increases by a factor of approximately 7.6. Time for the reaction to put on its running shoes and sprint towards the finish line!
To determine the factor by which the rate constant will increase when the temperature is raised, we can use the Arrhenius equation:
k = A * e^(-Ea/RT)
Where:
k is the rate constant
A is the pre-exponential factor
Ea is the activation energy in Joules per mole (J/mol)
R is the gas constant (8.314 J/(mol*K))
T is the temperature in Kelvin (K)
We can rewrite the equation as:
k2/k1 = (A * e^(-Ea/RT2))/(A * e^(-Ea/RT1))
Where:
k2 is the rate constant at the higher temperature
k1 is the rate constant at the lower temperature
T1 is the lower temperature (310K)
T2 is the higher temperature (325K)
Since we only need to find the factor by which the rate constant increases, we can cancel out the pre-exponential factor A:
k2/k1 = e^(-Ea/RT2)/e^(-Ea/RT1)
Now, we can substitute the given values into the equation:
k2/k1 = e^(-113000/(8.314*325))/e^(-113000/(8.314*310))
Calculating the exponents and dividing, we find:
k2/k1 ≈ 7.6
Therefore, the rate constant will increase by a factor of approximately 7.6 when the temperature is raised from 310K to 325K.
To determine the factor by which the rate constant will increase, you can use the Arrhenius equation:
k = A * e^(-Ea/RT)
where:
- k is the rate constant
- A is the pre-exponential factor
- Ea is the activation energy
- R is the gas constant (8.314 J/(mol*K))
- T is the absolute temperature in Kelvin
Given that the activation energy (Ea) is 113 kJ/mol, we need to convert it to joules per mole (J/mol) by multiplying it by 1000:
Ea = 113 kJ/mol * (1000 J/kJ) = 113000 J/mol
Now we can calculate the rate constant (k) at the initial temperature of 310K:
k1 = A * e^(-Ea/(R*310))
Next, we calculate the rate constant (k) at the higher temperature of 325K:
k2 = A * e^(-Ea/(R*325))
To find the factor by which the rate constant increases, we can divide k2 by k1:
factor = k2 / k1 = (A * e^(-Ea/(R*325))) / (A * e^(-Ea/(R*310)))
The pre-exponential factor (A) cancels out in the division, leaving:
factor = e^(-Ea/(R*325)) / e^(-Ea/(R*310))
Now we substitute the values:
factor = e^(-113000/(8.314*325)) / e^(-113000/(8.314*310))
Using a calculator, evaluate the exponential terms and divide them:
factor ≈ 7.6
Therefore, the rate constant will increase by a factor of approximately 7.6 when the temperature is raised from 310K to 325K.
Use the Arrhenius equation.
Solve for the ratio k2/k1. I obtained 7.56.