Question

A Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.50 LL of a 0.180 MM sucrose solution is allowed to react for 200 minmin .

Answer 1

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:

$\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}$

Molarity of sucrose solution = 0.120 M

Volume of solution = 2.65 L

$0.120mol/L=\frac{\text{Moles of sucrose solution}}{2.65L}\\\\\text{Moles of sucrose solution}=(0.120mol/L\times 2.65L)=0.318mol$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.8\times 10^{-4}s^{-1}$

t = time taken for decay process = 200 min = (200 × 60) = 12000 seconds

$[A_o]$ = initial amount of the sample = 0.318 moles

[A] = amount left after decay process = ?

Putting values in above equation, we get:

$1.8\times 10^{-4}=\frac{2.303}{12000}\log\frac{0.318}{[A]}$

$[A]=0.0367mol$

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of sucrose = 0.0367 moles

Molar mass of sucrose = 342.3 g/mol

Putting values in above equation, we get:

$0.0367mol=\frac{\text{Mass of sucrose}}{342.3g/mol}\\\\\text{Mass of sucrose}=(0.0367mol\times 342.3g/mol)=12.56g$

Hence, the mass of sucrose solution left is 12.56 grams