Eduqas A-Level Chemistry: Physical Chemistry overview quiz quiz

$+0.04\ \text{V}$

$-1.56\ \text{V}$

$+1.56\ \text{V}$

$+0.76\ \text{V}$

No reaction occurs because the two standard potentials are far too close together

$\text{I}_2$ oxidises $\text{Fe}^{2+}$ because it has the more positive electrode potential

Both species are reduced at the same time, so no electrons are actually transferred

$\text{Fe}^{3+}$ oxidises $\text{I}^-$ because it has the more positive potential

rate $= k[\text{A}][\text{B}]^2$

rate $= k[\text{A}][\text{B}]$

rate $= k[\text{A}]^2[\text{B}]$

rate $= k[\text{A}]^2[\text{B}]^2$

First order; it must therefore appear in the rate-determining step of the mechanism

Zero order; it reacts after the rate-determining step

Second order; this shows that the species is acting as a catalyst in the reaction

It cannot be a reactant at all if it does not appear in the rate equation

Larger ions with smaller charges

Lower temperature only

Smaller ions with larger charges

A larger molar mass

$2980\ \text{K}$

$553\ \text{K}$

$1610\ \text{K}$

$1106\ \text{K}$

It is thermodynamically feasible (spontaneous), though not necessarily fast

It will definitely happen quickly, since a negative value guarantees a high rate

It is endothermic, taking in heat energy from the surroundings as it proceeds

It has already reached equilibrium and so no further net change can take place

Increasing the pressure

Changing the temperature

Increasing the concentration of a reactant

Adding a catalyst

$1.00$

$2.87$

$4.74$

$5.00$

The ethanoic acid molecules react directly with the added acid to remove the extra $\text{H}^+$

Water in the solution neutralises the added acid by reacting with all of the extra $\text{H}^+$

The ethanoate ions react with the added $\text{H}^+$ to form ethanoic acid

The buffer increases its temperature, which drives off the added acid as a gas