The hydrolysis of sucrose into glucose and fructose in acidic water has a rate constant of at 25 degrees C.
Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.60 L of a 0.150 M sucrose solution is allowed to react for 190 minutes.
how do i work out this question? i am so lost :[
22.6 grams
To work out this question, you can follow these steps:
Step 1: Convert the given volume of solution into moles of sucrose.
- Volume of the solution = 2.60 L
- Concentration of sucrose (C) = 0.150 M
Number of moles of sucrose (n) = C * V
= 0.150 mol/L * 2.60 L
= 0.144 mol
Step 2: Use the given rate constant and time to calculate the fraction of sucrose hydrolyzed.
- Rate constant (k) =
- Time (t) = 190 minutes
Fraction of sucrose hydrolyzed = 1 - e^(-k*t)
Step 3: Calculate the mass of sucrose hydrolyzed.
- Molar mass of sucrose (M) = 342.30 g/mol (given)
Mass of sucrose (m) = n * M
= 0.144 mol * 342.30 g/mol
Hope this helps!
To solve this question, you can use the first-order rate equation to calculate the mass of sucrose that is hydrolyzed.
The first-order rate equation is given as:
Rate = k * [sucrose]
Where:
Rate is the rate of the reaction
k is the rate constant
[sucrose] is the concentration of sucrose
First, let's find the initial concentration of sucrose in moles:
0.150 M means 0.150 moles of sucrose per liter of solution.
So, the initial concentration of sucrose in moles is: 0.150 mol/L * 2.60 L = 0.39 moles
Now let's calculate the reaction rate using the given rate constant:
Rate = k * [sucrose]
Rate = k * 0.39
Next, we need to calculate the reaction time in seconds:
190 minutes * 60 seconds/minute = 11,400 seconds
Now, we'll use the integrated rate equation to find the amount of sucrose that has reacted after 190 minutes:
ln([sucrose]t/[sucrose]0) = -kt
Where:
[sucrose]t is the concentration of sucrose at time t (which is what we're trying to find)
[sucrose]0 is the initial concentration of sucrose
k is the rate constant
t is the reaction time in seconds
Since the reaction is first-order, we can rewrite the equation as:
ln([sucrose]t/0.39) = -kt
Plugging in the values we know:
ln([sucrose]t/0.39) = -k * 11,400
Now we need to rearrange the equation to solve for [sucrose]t:
[sucrose]t = 0.39 * e^(-k * 11,400)
Finally, substitute the given rate constant ( ) and calculate [sucrose]t.
I hope this explanation helps you in solving the question!
First you get the rate constant that you didn't bother to type in. Then I would convert M and L to moles and use
ln(No/N) = kt