Question

The hydrolysis of sucrose was studied with the help of polarimeter and following data were collected time (min.) \(\begin{array}{lll}: 0 & 70 & \infty\end{array}\) observed rotation (degrees) \(\begin{array}{lll}: 44 & 16.5 & -11\end{array}\) when the reaction mixture will be optically inactive ? (Given: \(\ln 2=0.7, \ln 3=1.1, \ln 5=1.6\) ) (a) 16 min. (b) \(69.47 \mathrm{~min}\). (c) \(160 \mathrm{~min}\). (d) none of these

Short answer

To find the time when the reaction will be optically inactive, first calculate the rate constant 'k' using the given data points. Then use 'k' to determine the time 't' for optical inactivity. However, as per available data and the logarithmic nature of the decay, the exact moment when the rotation is zero cannot be pinpointed; therefore, the answer is (d) none of these, given that none of the provided choices exactly match the point at which the optical activity ceases.

Step-by-step solution

Understand the Initial Conditions

Start by examining the data provided. The given data shows the rotation of polarized light through a sucrose solution at different times during the hydrolysis reaction. Initially, at time 0 min, the observed rotation is 44 degrees. It means that the sucrose was optically active because it rotated the plane of polarized light.

Determine the Conditions for Optical Inactivity

As the reaction progresses, hydrolysis of sucrose will produce glucose and fructose, which will affect the optical activity. The reaction mixture will become optically inactive when the optical rotations due to glucose and fructose cancel each other out. This is the point at which the observed rotation becomes zero.

Find the Relationship Between Optical Rotation and Time

The problem suggests that optical rotation '[R]' depends on time 't', likely following first-order reaction kinetics, where [R] decreases exponentially with time. If that assumption holds, we can use first-order kinetics represented as \[ \ln\left(\frac{[R]_0}{[R]_t}\right) = kt \] where \([R]_0\) is the initial rotation, \([R]_t\) is the rotation at time 't', 'k' is the rate constant, and 't' is time.

Calculate the Rate Constant 'k'

Using the data points provided (time: 0 min and 70 min; rotation: 44 degrees and 16.5 degrees), we can establish the rate constant 'k' using the logarithmic formula. We can substitute into the formula as follows: \[ \ln\left(\frac{44}{16.5}\right) = kt \] Using the data for 't' = 70 min, we calculate 'k'.

Solve for the Rate Constant 'k'

Now calculate the value of 'k' by solving the equation with the given logarithmic values: \[ \ln\left(\frac{44}{16.5}\right) = k \times 70 \] \[ \ln\left(\frac{44}{16.5}\right) = \ln\left(\frac{88}{33}\right) = \ln\left(\frac{8 \times 11}{3 \times 11}\right) = \ln\left(\frac{8}{3}\right) \] Given that \(\ln 2 = 0.7\) and \(\ln 3 = 1.1\), we have: \[ \ln\left(\frac{8}{3}\right) = \ln(8) - \ln(3) = 3\ln(2) - \ln(3) = 3 \times 0.7 - 1.1 \] Finally calculate 'k' by dividing this value by 70.

Calculate Time 't' for Optical Inactivity

Using the calculated rate constant 'k', determine the time 't' at which the observed rotation would be 0 degrees (optically inactive). The equation is again: \[ \ln\left(\frac{[R]_0}{0}\right) = kt \] This equation is actually undefined, since division by zero is not possible. However, because we are looking for when the rotation is zero, we need the numerator (initial rotation) to have decreased to zero effectively. This would require the logarithm to be very large, effectively equating kt to infinity.

Solve for Time 't' When Rotation Is Zero

Since the logarithm of a number that approaches zero goes to negative infinity, we aim to find the time 't' when the rotation is practically zero (indicating the time taken for the rotation to go from 44 degrees to almost zero). Use the previously calculated k and solve for 't'.

Analyze the Data for the Infinite Time Condition

The data also includes an observed rotation at infinite time, which is -11 degrees, suggesting that fructose contributes a negative rotation that becomes evident once sucrose is fully hydrolyzed. This is a clue that the time when the rotation is zero must be before infinite time and after the last measured time (70 min).

Choose the Correct Answer

Given the calculated rate constant 'k' and the data provided, we can now compare the available options for the time at which the rotation is zero. If none of the options match our calculated time, the correct answer would be none of these.

Key concepts

Optical Activity

Optical activity is a property of certain substances to rotate the plane of polarized light. This occurs due to the asymmetric structure of molecules, which can be left- or right-handed, referred to as 'chirality'. Substances that rotate polarized light to the right are called 'dextrorotatory', while those that rotate it to the left are 'levorotatory'. In the context of the hydrolysis of sucrose, which is a dextrorotatory molecule, the process yields two products: glucose and fructose. Glucose is also dextrorotatory, but fructose is levorotatory and has a stronger optical activity in the opposite direction. This means that as sucrose is converted into glucose and fructose, the solution's overall optical activity decreases, eventually reaching a point where it can be optically inactive when the contributions from glucose and fructose cancel each other out.

Polarimetry

Polarimetry is the technique used to measure the angle of rotation caused by passing polarized light through an optically active substance. It involves using a device known as a polarimeter. In our exercise, the polarimeter is used to track the change in optical rotation as sucrose undergoes hydrolysis. At time zero, the observed rotation was 44 degrees. During the reaction, this angle changes as the concentration of the optically active molecules changes. By monitoring these changes over time, we can gain insights into the reaction's progress and eventually determine when the reaction mixture becomes optically inactive, which is a key point in understanding reaction kinetics.

First-Order Reaction Kinetics

In first-order reaction kinetics, the rate at which a reactant is consumed is directly proportional to its concentration at any given moment. Mathematically, this relationship is often represented using the equation \[ \[ ln\left(\frac{{[R]_0}}{{[R]_t}}\right) = kt \] \] where \( [R]_0 \) is the initial concentration (or in the case of our exercise, initial rotation), \( [R]_t \) is the concentration (rotation) at time \( t \) , \( k \) is the rate constant, and \( t \) is the time. This equation suggests that the logarithm of the ratio of initial to remaining rotation is proportionate to time, and indicates an exponential decrease over time. When this principle is applied to the hydrolysis of sucrose, it allows us to predict when exactly the reaction mixture will become optically inactive.

Chemical Equilibrium

Chemical equilibrium refers to the state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, causing the concentrations of the reactants and products to remain constant over time. It doesn't apply directly in the context of this exercise, because the hydrolysis of sucrose is typically a one-way (irreversible) reaction. However, the concept of equilibrium is conceptually important when considering that the reaction will reach a point where the amounts of glucose and fructose produced yield no net optical rotation—this is a form of equilibrium between the optical activities of the reaction's products. Understanding this helps us recognize that the hydrolysis reaction will change the optical rotation until it reaches this neutral point even though the reaction itself does not reach a true chemical equilibrium.