Explanation​:

The theoretical yield is the maximum amount of product that will be formed if the limiting reagent is consumed completely.

Explanation:
The rate expression can only be determined experimentally.

Explanation​:
Units for $k$ are determined by dividing the units of rate by the units of the concentrations raised to their respective powers.
For a 3rd order reaction, k=mol dm−3 s−1/mol3 dm−9 = mol−2 dm6 s−1.

Explanation​:
Increasing the volume decreases the pressure which shifts the position of the equilibrium to the right where there are more moles of gas.
The forward reaction is endothermic and so increasing the temperature favors this reaction.
A catalyst affects both the rate of the forward and reverse reactions equally and has no effect on the equilibrium yield.

Explanation​:
Rate is directly proportional to the concentration of $M$ as it is first order. As the concentration of $M$ has been multiplied by 1.5, the rate (0.5 s−1) must also be multiplied by 1.5.

Explanation:​
Students should be aware of the definition of frequency factor. However, questions on this topic are somewhat uncommon, and students could confuse this concept with the idea that molecules capable of successful collision lie above Ea​ in a Boltzmann distribution (leading them to select option D). The difficulty level is therefore at the higher end of the “easy” range.

Explanation​:
Graph D shows that rate is proportional to [A]2.
A and B are concentration-time graphs. C is a first order rate-concentration graph as they are directly proportional

Explanation​:
A catalyst provides an alternative pathway for the reaction lowering the activation energy. It affects both the forward and reverse reaction rates equally. The position of equilibrium will not change.

Explanation:​
All graphs are [A] vs time.
A linear plot of [A] vs time represents a zero order reaction, B and D represent first order reactions (B represents [A] as a product and is therefore increasing).
C is a second order plot of [A] vs time and has a steeper initial rate (slope) when compared to first order reactions.

Explanation:​
The rate constant, $k$, is the feature of the rate equation that is affected by temperature according to the Arrhenius equation k=Ae−Ea/RT.
The graph of rate vs temperature always shows this general trend, and similar questions have been asked several times in past papers.
The question is of moderate difficulty as the use of a specific example and reference to rate rather than k may derail some students. Confusion with the Arrhenius plot could also lead students to select option B.

Explanation​:
Both sides of the equation have the same number of gaseous moles, so the equilibrium position is unaffected by changes in pressure.

Explanation:​
The number of molecules remains constant. Temperature only changes the particles’ kinetic energy; when temperature increases, the kinetic energy of the particles also increases.

Explanation​:
OH− ions react with the H+ ions, thus decreasing their concentration. The equilibrium shifts towards the reactants to produce more H+ and resist this change (Le Chatelier’s principle). This produces more yellow chromate ions.
The value of Kc is only affected by temperature so this is unchanged.

Explanation​:
Using the Arrhenius equation k = Ae−Ea/RT, a linear relationship will be created by graphing ln k vs. 1/T​

Explanation:​
Kc>1, and so lies towards the right with the yellow product at equilibrium; the color would therefore be a pale green due to more yellow than blue.
When an acid is added, it reacts with the OH− reducing its concentration. Equilibrium shifts to the right to oppose that change.

Kc only changes with temperature. Adding an alkali would increase the OH− concentration and therefore the equilibrium would shift to the left to oppose that change. Kc​ again would remain the same as it is only affected by temperature.

Explanation:
The balanced chemical equation is 2C3H7OH(l)+9 O2(g)→6 CO2(g)+8 H2O(l). The sum of the coefficients is $25$.

Explanation:
Using trials 1 and 2, as [X] doubles, rate also doubles therefore first order with respect to [X].
Using trials 2 and 3, as [Y] halves, rate stays the same and so, zero order with respect to [Y].
This gives a slow step that depends on unimolecular X, and not Y.

Explanation​:
The equilibrium shifts to favor the endothermic reaction (forward) when the temperature is increased to resist that change (Le Chatelier’s principle).
Temperature is the only factor that changes the value of the equilibrium constant.