Respuesta :
The question is incomplete, here is the complete question:
The hydrolysis of sucrose into glucose and fructose in acidic water has a rate constant of 1.8×10⁻⁴ s⁻¹ at 25°C. Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.65 L of a 0.120 M sucrose solution is allowed to react for 200 min.
Answer: The mass of sucrose solution left is 12.56 grams
Explanation:
Molarity is calculated by using the equation:
[tex]\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}[/tex]
Molarity of sucrose solution = 0.120 M
Volume of solution = 2.65 L
[tex]0.120mol/L=\frac{\text{Moles of sucrose solution}}{2.65L}\\\\\text{Moles of sucrose solution}=(0.120mol/L\times 2.65L)=0.318mol[/tex]
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = [tex]1.8\times 10^{-4}s^{-1}[/tex]
t = time taken for decay process = 200 min = (200 × 60) = 12000 seconds
[tex][A_o][/tex] = initial amount of the sample = 0.318 moles
[A] = amount left after decay process = ?
Putting values in above equation, we get:
[tex]1.8\times 10^{-4}=\frac{2.303}{12000}\log\frac{0.318}{[A]}[/tex]
[tex][A]=0.0367mol[/tex]
To calculate the mass from given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of sucrose = 0.0367 moles
Molar mass of sucrose = 342.3 g/mol
Putting values in above equation, we get:
[tex]0.0367mol=\frac{\text{Mass of sucrose}}{342.3g/mol}\\\\\text{Mass of sucrose}=(0.0367mol\times 342.3g/mol)=12.56g[/tex]
Hence, the mass of sucrose solution left is 12.56 grams
