Respuesta :
Answer:
Mass of sucrose that is hydrolyzed = 123.55 g
Explanation:
Since the reaction is a first ofree reaction,
Let the initial concentration of sucrose be C₀
And the concentration of sucrose left at any time be C.
r = (dC/dt) = -kC (Minus sign because it's a rate of reduction)
K = rate constant
(dC/dt) = - kC
(dC/C) = -kdt
∫ (dC/C) = -k ∫ dt
Solving the two sides as definite integrals by integrating the left hand side from C₀ to C and the Right hand side from 0 to t.
We obtain
In (C/C₀) = -kt
(C/C₀) = (e⁻ᵏᵗ)
C = C₀ e⁻ᵏᵗ
So, we can find the concentration of sucrose left at the given time
k = 1.8 × 10⁻⁴ s⁻¹
t = 200 minutes = 200 × 60 = 12000 s
kt = 12000 × 1.8 × 10⁻⁴ = 2.16
C₀ = 0.160 M
C = 0.160 e⁻²•¹⁶ = 0.01845 M
So, the concentration that has been used up = 0.160 - 0.01845 = 0.142 M
To calculate the mass of sucrose that has reacted, we'll convert the concentration used up into number of moles.
Concentration = (number of moles)/(volume in Litres)
number of moles = concentration × volume in Litres
Volume in Litres = 2.55L
Number of moles = 0.142 × 2.55 = 0.361 moles
Then, mass = number of moles × Molar mass
Molar mass of sucrose = 342.3 g/mol
Mass of sucrose used up = 0.361 × 342.3 = 123.55 g
The mass of sucrose is 16 g.
Using the equation;
ln[A] = ln[Ao] - kt
were;
[A] = concentration at time t
[Ao] = initial concentration
k = rate constant
t = time taken
Substituting values;
ln[A] = ln(0.160 M) - [ 1.8×10−4s−1 × 200 × 60]
[A] = e^ln(0.160 M) - [ 1.8×10−4s−1 × 200 × 60]
[A] = e^-1.83 - 2.16
[A] = e^-3.99
[A] = 0.0185 M
Now;
number of moles = concentration × volume
mass/342 g/mol = 0.0185 M × 2.55 L
Mass = 0.0185 M × 2.55 L × 342 g/mol
Mass = 16 g
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