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 $ln(2)$ term appears because k1 is double k2, and the factor of 1000 appears in the denominator to convert J mol−1to kJ mol−1.
Option B is almost correct, but the temperature terms 1/298 and 1/308 would need to be swapped to give a positive value for the activation energy.
The question is in the “hard” or “very hard” category since students must be able to identify the correct terms in the question for use in the equation, rearrange the equation correctly, and also perform a unit conversion.

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:
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.

Therefore, 0.5 moles of H2require 0.167 moles of N2 to completely react. There are $s$ of N2NX2 available so N2 is the reactant in excess.

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

Explanation:
Q>Kc, therefore the backward reaction will be favored as the system approaches dynamic equilibrium.
The backward reaction is endothermic, so the temperature will decrease.
There are more moles of gas on the left and so pressure increases. NO2 is red-brown and so the gas mixture will get darker.

Explanation:
A dynamic equilibrium is reached in a closed system when the rates of the forward and reverse reactions are equal. The concentrations of reactants and products do not change but are non-stop being interconverted. The equilibrium could favor one side over the other.

Explanation:

The correct balanced equation is:

2LiOH + H2SO4 → Li2SO4 + 2H2O.

2+1+1+2 = 6

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:
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:

Kc is written as the concentrations of the products divided by the concentrations of the reactants each raised to the power of the coefficient in the balanced equation.

Explanation:
The molecularity is given by the power to which the species is raised to in the rate expression. X is raised to the power of two meaning the rate determining step involves two X molecules.

Explanation:
The ratio of SO3:SO2:O2 is 2:2:1. Therefore, half the number of moles of oxygen would be produced compared to SO2

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.

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:
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.