Chemical Engineering - MCQ Practice Questions

Understanding:

We must find the theoretical oxygen demand for complete oxidation of a glucose solution.

Formula:

The complete oxidation reaction for glucose is:

ThOD is the mass of ${\text{O}}_2$ required per unit mass of compound:

Step 1: Determine moles of ${\text{O}}_2$ per mole of glucose

From the balanced equation, 1 mole of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ requires 6 moles of ${\text{O}}_2$.

Step 2: Calculate the ThOD ratio

Step 3: Calculate ThOD for the given concentration

Verification:

Molecular weight of $6{\text{O}}_2=6\times 32=192\text{g/mol}$. Molecular weight of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6=180\text{g/mol}$. For $180\text{mg/L}$ glucose (which is exactly 1 mmol/L), ThOD $=6\times 32=192\text{mg/L}$.

Answer:

The theoretical oxygen demand of a $180\text{mg/L}$ glucose solution is $192\text{mg/L}$.

Quick Tip:

Glucose is a convenient benchmark: its molecular weight is $180\text{g/mol}$ and its ThOD ratio is $192/180\approx 1.067$. Because glucose is readily biodegradable, its BOD${}_5$/ThOD ratio approaches $0.92$–$0.95$.

Understanding:

We need to find the number-average molecular weight given the degree of polymerization and the molecular weight of the repeating unit.

Formula:

Step 1: Substitute values

Answer:

The number-average molecular weight of the polymer is $5.20\times 10^5\text{g/mol}$.

Quick Tip:

This formula applies when a single repeating unit mass is given. Note that $M_0=104\text{g/mol}$ corresponds to polystyrene (${\text{C}}_8{\text{H}}_8$, MW = 104), a common example in polymer problems.

Understanding:

We must calculate the kinetic chain length from the rates of propagation and initiation.

Formula:

Step 1: Divide the two rates

Answer:

The kinetic chain length is 500.

Quick Tip:

The kinetic chain length equals the average number of monomer units added per active chain end before termination. For termination by combination, the degree of polymerization $\text{DP}=2\bar{\nu }$; for disproportionation, $\text{DP}=\bar{\nu }$.

Understanding:

We need to identify which polymer is formed via condensation polymerization among the given options.

Step 1: Classify each polymer by mechanism

Polyethylene (PE), polyvinyl chloride (PVC), and polystyrene (PS) are all formed by addition (chain-growth) polymerization of vinyl monomers. No small molecule is eliminated during their formation.

Step 2: Identify the condensation polymer

Nylon-6,6 is produced by the reaction between hexamethylenediamine (${\text{H}}_2\text{N}-({\text{CH}}_2{)}_6-{\text{NH}}_2$) and adipic acid ($\text{HOOC}-({\text{CH}}_2{)}_4-\text{COOH}$). A water molecule (${\text{H}}_2\text{O}$) is eliminated at each step, making it a classic condensation (step-growth) polymer.

Answer:

Nylon-6,6 is produced by condensation (step-growth) polymerization.

Quick Tip:

A quick test: if the monomer has two different functional groups (e.g., $-{\text{NH}}_2$ and $-\text{COOH}$) and a small molecule like water or HCl is released, it is condensation polymerization.

Understanding:

We must calculate the polydispersity index from the weight-average and number-average molecular weights.

Formula:

Step 1: Substitute and simplify

Answer:

The polydispersity index of the polymer sample is 1.50.

Quick Tip:

PDI = 1 for a perfectly monodisperse polymer. Living polymerization techniques (e.g., ATRP, RAFT) achieve PDI close to 1, whereas conventional free-radical polymerization gives PDI in the range 1.5–2.0.

Understanding:

We need to identify the critical condition on the Flory-Huggins interaction parameter $\chi$ for complete miscibility of a polymer solution.

Step 1: State the Flory-Huggins criterion

In Flory-Huggins theory, the critical value of the interaction parameter for a high-molecular-weight polymer (as $n\to \infty$, where $n$ is the degree of polymerization) is:

Step 2: Apply the limit for large $n$

As the chain length $n\to \infty$:

Step 3: State the stability condition

For the polymer to be completely miscible with the solvent at all compositions, the interaction parameter must be less than the critical value:

When $\chi >0.5$, phase separation occurs. When $\chi =0.5$, this defines the theta ($\Theta$) condition.

Answer:

For thermodynamic stability (miscibility) of a high-molecular-weight polymer solution, $\chi$ must be less than 0.5.

Quick Tip:

The condition $\chi =0.5$ corresponds to the theta ($\Theta$) solvent condition, where polymer-solvent and polymer-polymer interactions exactly balance, and the chain behaves ideally.

Understanding:

We must compute the number-average molecular weight from the given molar fractions.

Formula:

Step 1: Calculate the numerator $\sum N_i{M}_i$

Step 2: Calculate the denominator $\sum N_i$

Step 3: Compute ${\bar{M}}_n$

Answer:

The number-average molecular weight is approximately $18,333\text{g/mol}$.

Quick Tip:

${\bar{M}}_n$ is sensitive to the presence of low-molecular-weight species (it weights all chains equally), while ${\bar{M}}_w$ is more sensitive to high-molecular-weight chains.

Understanding:

We need to identify the correct physical description of the glass transition temperature $T_g$ for an amorphous polymer.

Step 1: Define $T_g$

The glass transition temperature $T_g$ is a characteristic of amorphous (non-crystalline) polymers and the amorphous regions of semi-crystalline polymers. Above $T_g$, the polymer chains have sufficient thermal energy for large-scale cooperative segmental motion, giving a rubbery or viscous behaviour. Below $T_g$, this motion is frozen out and the polymer is hard, brittle, and glassy.

Step 2: Eliminate incorrect options

Answer:

The glass transition temperature is the temperature at which large-scale segmental motion of the polymer backbone ceases upon cooling, making the polymer rigid and glassy.

Quick Tip:

For most polymers, $T_g\approx \frac{2}{3}{T}_m$ (in Kelvin), a useful empirical rule for quick estimation when only $T_m$ is known.

Understanding:

We need to recall the Smith-Ewart Case II kinetics condition for emulsion polymerization and the resulting average number of radicals per particle.

Step 1: State the Smith-Ewart cases

Smith-Ewart theory classifies emulsion polymerization into three cases based on the average number of radicals per particle, $\bar{n}$:

Step 2: Identify Case II

In Case II (the classic Smith-Ewart condition), a particle alternates between having one radical (active, growing) and zero radicals (dormant). Because termination is instantaneous when a second radical enters, the time-averaged number of radicals per particle is exactly:

Answer:

For Smith-Ewart Case II kinetics, the average number of radicals per particle is 0.5.

Quick Tip:

The high rate of emulsion polymerization despite $\bar{n}=0.5$ arises because there are a very large number of particles (typically $10^{14}$–$10^{15}$ per litre), not because each particle has many radicals.

Understanding:

We must find the extent of reaction $p$ required to achieve a given degree of polymerization in step-growth polymerization.

Formula:

The Carothers equation for a stoichiometrically balanced bifunctional system is:

Step 1: Rearrange for $p$

Step 2: Substitute ${\bar{X}}_n=200$

Verification:

Answer:

An extent of reaction of $p=0.995$ (99.5% conversion) is needed to achieve ${\bar{X}}_n=200$.

Quick Tip:

This result highlights a critical feature of step-growth polymerization: extremely high conversions (>99%) are required to produce high-molecular-weight polymers. This is why precise stoichiometry and removal of the condensate are essential in industrial nylon or polyester synthesis.