The rate equation for the hydrolysis of sucrose to fructose and glucose is first order. After 27 minutes the sucrose concentration decreased from 0.0146M to 0.0132M. Find the rate constant K.
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To find the rate constant (K) for the hydrolysis of sucrose, we can use the first-order rate equation:
ln(A/A0) = -Kt
Where A is the concentration at time (t), A0 is the initial concentration, K is the rate constant, and t is the time.
In this case, we are given the initial concentration (A0) as 0.0146M, the concentration after 27 minutes (A) as 0.0132M, and the time (t) as 27 minutes.
Now we can substitute these values into the equation and rearrange it to solve for K:
ln(A/A0) = -Kt
ln(0.0132M/0.0146M) = -K * 27 minutes
Taking the natural logarithm of the concentration ratio gives:
ln(0.0132M/0.0146M) = -K * 27 minutes
Simplifying the right side of the equation:
ln(0.0132M/0.0146M) = -27K
Now we can solve for the rate constant (K):
K = -ln(0.0132M/0.0146M) / 27 minutes
Using a calculator, we can evaluate this expression:
K ≈ -ln(0.0132M/0.0146M) / 27 minutes
K ≈ -0.0913 / 27 minutes
K ≈ -0.003382
Therefore, the rate constant (K) for the hydrolysis of sucrose is approximately -0.003382 min^-1.
To find the rate constant (K) for a first-order reaction, you can use the following equation:
ln([A]t/[A]0) = -Kt
where:
[A]t is the concentration of reactant A at time t
[A]0 is the initial concentration of reactant A
K is the rate constant
t is the time
In this case, we have the following information:
[A]t = 0.0132 M (concentration after 27 minutes)
[A]0 = 0.0146 M (initial concentration)
t = 27 minutes
Using these values, we can rearrange the equation to solve for K:
ln([A]t/[A]0) = -Kt
ln(0.0132/0.0146) = -K * 27
Now, let's solve for K:
ln(0.0132/0.0146) = -K * 27
Simplify the left side:
ln(0.0132/0.0146) ≈ -K * 27
Calculate the natural logarithm:
-0.089 <-K * 27
Divide both sides by -27:
-0.089/-27 ≈ K
Simplify:
0.0033 ≈ K
Therefore, the rate constant (K) for the hydrolysis of sucrose is approximately 0.0033.
For first order,
ln(No/N) = kt
No = given
N = given
t = given
Plug and chug.