Surfaces, Interfaces and Colloids Questions and Answers

Answer: C

At fixed coverage, \(K \propto P^{-1}\).

For non-dissociative adsorption the Langmuir isotherm gives the fractional coverage as

\[ \theta = \frac{KP}{1 + KP} \]

If \(\theta\) is held fixed, the product \(KP\) must be fixed, \(KP = \theta/(1-\theta)\), so K is inversely proportional to P.

\(K \propto P\) inverts the relation. The square-root forms belong to dissociative adsorption, where \(\theta\) depends on \((KP)^{1/2}\), which the question rules out.

Answer: B

The micelle concentration is 0.001 M.

Above the critical micelle concentration, surfactant in excess of the CMC goes into micelles, while the CMC amount stays as free monomer:

\[ \begin{aligned} [\text{surfactant in micelles}] &= 0.088 - 0.008 = 0.080\ \text{M} \\ [\text{micelles}] &= \frac{0.080}{80} = 0.001\ \text{M} \end{aligned} \]

Dividing the full 0.088 M by 80 gives 0.0011 M, close but not the intended value; 0.008 M is the CMC itself and 0.088 M the total concentration, neither of which is a micelle concentration.

Answer: D

The Freundlich isotherm is plot (D).

The isotherm is \(x/m = kP^{1/n}\). Taking logarithms gives a straight line:

\[ \log\frac{x}{m} = \log k + \frac{1}{n}\log P \]

The line has a positive slope \(1/n\) and a positive intercept \(\log k\), which is what (D) shows. (A) levels off to a plateau, the shape of a Langmuir isotherm as the surface saturates, which is not a log-log line. (B) is a line through the origin that turns upward, and (C) has a negative slope, so neither fits the equation.

Answer: D

The untrue statement is It catalyses the non-spontaneous reactions.

A catalyst lowers the activation energy, providing a faster route, but it does not change \(\Delta G\) or the equilibrium constant. It can therefore only speed up reactions that are already thermodynamically feasible; it cannot make a non-spontaneous reaction go.

The other statements are true: catalysts speed up spontaneous reactions, a small amount suffices because it is regenerated, and it leaves both the Gibbs energy change and \(K\) unchanged.

Answer: B

Sodium lauryl sulphate is an anionic detergent.

Its formula is \(\mathrm{CH_3(CH_2)_{11}OSO_3^-\,Na^+}\). In water the sodium ion separates and the long-chain sulphate ion, which carries the negative charge, does the cleaning, so it is anionic. It is common in shampoos and toothpastes.

Cationic detergents, such as cetyltrimethylammonium bromide, have a positively charged head; non-ionic detergents, such as polyethylene glycol esters, carry no charge.

Answer: C

The correct plot is C, a rise to a maximum followed by a fall.

When both reactants compete for the same sites (the Langmuir-Hinshelwood mechanism) the rate is proportional to the product of their fractional coverages:

\[ \begin{aligned} r &= \frac{k K_{A}K_{B} p_{A} p_{B}}{D^{2}} \\ D &= 1 + K_{A}p_{A} + K_{B}p_{B} \end{aligned} \]

At constant ethylene pressure the rate first rises with p_H2, because more hydrogen is adsorbed. At high p_H2 the denominator grows as p², hydrogen crowds the alkene off the surface and the rate falls. A constant rate (A), a saturation curve (B) and a straight line (D) describe zero-order, single-site or first-order behaviour, not competitive adsorption. The axis labels print C₂H₆ for ethylene.

Answer: B

The correct statement is that at least one reactant must be chemisorbed on the catalyst.

A heterogeneous catalyst works by forming chemical bonds with an adsorbed reactant, which weakens its bonds and lowers the activation energy. Physisorption, held by weak van der Waals forces, does not change the molecule enough to activate it, so it is not required; chemisorption of at least one reactant is.

The number of active sites increases, not decreases, with surface area, which is why catalysts are finely divided or dispersed on supports. Surface area is measured by gas adsorption, as in the BET method with nitrogen, not by spectroscopy.

Answer: A

The kinetic energy of the photoelectrons from X (Li metal) will be higher than from Y (Li₂O).

In XPS the measured kinetic energy is

\[ KE = h\nu - BE - \phi \]

In Li₂O lithium is present as Li⁺. The loss of electron density makes the remaining 1s electrons more tightly held, so the Li 1s binding energy is higher in the oxide than in the metal (a chemical shift). For the same X-ray energy, a higher binding energy means a lower kinetic energy.

So Y gives photoelectrons of lower kinetic energy, ruling out D, and the binding energy for Y is higher, not lower, ruling out C. Option B compares Y with itself.

Answer: B

The values imply (ii) and (iii).

Surface tension reflects the strength of the forces holding the liquid together. Bromine, held by van der Waals forces, has the lowest value (41.0 mN m⁻¹); water, held by hydrogen bonds, is higher (72.0); and mercury, held by metallic bonding, is far higher (486.2).

So hydrogen bonding is stronger than van der Waals bonding (ii), and metallic bonding is stronger than hydrogen bonding (iii). Statement (iv) is contradicted by the twelvefold range of values, and statement (i) is not supported, since no liquid here is held by covalent bonds between molecules and mercury's value is the highest.