IMO Practice Test — Surface Chemistry
13 Questions • 15 min • Olympiad level
15:00
Question 1 of 13
For a given adsorbate at constant temperature, doubling the pressure increases $\frac{x}{m}$ by a factor of $\sqrt{2}$. The value of $n$ in the Freundlich isotherm is:
$2$
$1$
$\frac{1}{2}$
$4$
Explanation: If $\frac{x}{m}\propto p^{1/n}$ and a factor-2 pressure gives factor $2^{1/n}=2^{1/2}$, then $\frac{1}{n}=\frac{1}{2}$, so $n=2$.
Question 2 of 13
On a $\log\frac{x}{m}$ vs $\log p$ plot the slope is found to be $0.25$. The value of $n$ is:
$0.25$
$4$
$0.75$
$2$
Explanation: Slope $=\frac{1}{n}=0.25$, so $n=4$.
Question 3 of 13
Which of the following increases when physisorption changes into chemisorption as temperature rises?
reversibility
number of layers
enthalpy of adsorption
van der Waals character
Explanation: Chemisorption involves chemical bonds, so the enthalpy of adsorption increases sharply compared with physisorption.
Question 4 of 13
A catalyst lowers the activation energy of a reaction from $75$ to $50\ \text{kJ mol}^{-1}$. The equilibrium constant of the reaction:
increases
decreases
becomes zero
remains unchanged
Explanation: A catalyst lowers $E_a$ for forward and reverse steps equally, so $K$ (and $\Delta G$) is unchanged.
Question 5 of 13
Among $\text{Na}_3\text{PO}_4$, $\text{Na}_2\text{SO}_4$ and $\text{NaCl}$, the strongest coagulant for a positively charged $\text{Fe(OH)}_3$ sol is:
$\text{Na}_3\text{PO}_4$
$\text{Na}_2\text{SO}_4$
$\text{NaCl}$
all are equal
Explanation: A positive sol is coagulated by anions; the highest-charge anion $\text{PO}_4^{3-}$ (from $\text{Na}_3\text{PO}_4$) is most effective.
Question 6 of 13
The Brownian motion of colloidal particles is mainly responsible for:
their colour
their stability against settling
their charge
the Tyndall effect
Explanation: Continuous random motion keeps particles suspended and prevents them from settling under gravity.
Question 7 of 13
CO and $\text{H}_2$ over a $\text{Cu}/\text{ZnO}/\text{Cr}_2\text{O}_3$ catalyst give mainly:
methane
higher hydrocarbons
methanol
carbon and water
Explanation: This illustrates selectivity: that catalyst directs CO + $\text{H}_2$ to methanol (Ni gives methane, Co gives hydrocarbons).
Question 8 of 13
The charge on a colloidal sol arises mainly from:
Brownian motion
the Tyndall effect
preferential adsorption of ions
gravity
Explanation: Colloidal particles preferentially adsorb common ions from the medium, acquiring charge (the electrical double layer).
Question 9 of 13
Which is NOT a feature of enzyme catalysis?
high specificity
optimum pH
optimum temperature
activity independent of temperature
Explanation: Enzyme activity depends strongly on temperature, peaking near an optimum and falling sharply on denaturation.
Question 10 of 13
In the Cottrell electrostatic precipitator the smoke particles are removed by:
heating to high temperature
neutralising their charge so they coagulate and settle
dissolving them in water
the Tyndall effect
Explanation: Charged smoke (carbon) particles are neutralised at a charged electrode, then coagulate and fall, cleaning the gas.
Question 11 of 13
A sol of $\text{As}_2\text{S}_3$ is negatively charged. The minimum amount needed for coagulation (the coagulation value) is smallest for:
$\text{NaCl}$
$\text{MgCl}_2$
$\text{AlCl}_3$
$\text{KCl}$
Explanation: Higher cation charge means greater coagulating power and a smaller coagulation value, so $\text{AlCl}_3$ ($\text{Al}^{3+}$) needs the least.
Question 12 of 13
Formation of a river delta where river water meets sea water is due to:
the Tyndall effect
peptisation of clay
coagulation of colloidal clay by electrolytes in sea water
electrophoresis
Explanation: Electrolytes in sea water coagulate the colloidal clay carried by the river, which deposits as a delta.
Question 13 of 13
For physisorption of a gas on a solid, a graph of amount adsorbed against temperature (at fixed pressure) shows the amount:
rising continuously
falling as temperature rises
constant
rising then constant
Explanation: Physisorption is exothermic, so the amount adsorbed decreases as temperature increases.