NESA Chemistry Static and Dynamic Equilibrium

If the rate of evaporation were less than the rate of condensation, the amount of liquid water would increase, causing the water level to rise.

If the rate of evaporation were greater than the rate of condensation, the amount of liquid water would decrease, causing the water level to drop.

While the rates are equal, they are not zero because molecules are constantly escaping and returning to the liquid surface in a state of dynamic equilibrium.

This is the total energy absorbed to produce one full mole of glucose. Producing one mole of glucose requires 6 moles of $CO_2$, not just one.

Photosynthesis is an endothermic process that absorbs energy from sunlight. This option incorrectly states that energy is released.

This describes the energy released during the combustion of one mole of glucose, rather than the energy absorbed per mole of $CO_2$ during photosynthesis.

The enthalpy change (whether a reaction is exothermic or endothermic) determines the thermodynamics of the reaction, not its kinetics or rate. Highly exothermic reactions can still be very slow.

Reactions requiring the simultaneous collision of many types of particles are statistically less likely to occur, which would typically result in a slower reaction rate, not a faster one.

The prompt explicitly states the reaction is rapid at room temperature, indicating that an external heat source like a flame is not required to overcome the activation energy.

This is a true characteristic of dynamic equilibrium. The system is "dynamic" precisely because both the forward and reverse reactions continue to occur.

This is a true characteristic. The fundamental definition of dynamic equilibrium is that the forward and reverse reaction rates are exactly equal.

This is a true characteristic. According to Le Chatelier's principle, changing the temperature of a system at equilibrium will shift its position to favor either the endothermic or exothermic direction.

While combustion is highly exothermic, exothermicity alone does not make a reaction irreversible. Many reversible reactions have exothermic forward reactions.

In a combustion reaction, the products ($\text{CO}_2$ and $\text{H}_2\text{O}$) are actually more stable than the reactants, which is why the reaction releases energy.

While adding a reactant like oxygen favors the forward reaction according to Le Chatelier's principle, this describes shifting an equilibrium rather than defining a reaction as irreversible.

A catalyst does not change the speed of the particles. Increasing the temperature of the system would increase particle speed.

A catalyst does not increase the frequency of collisions. Factors like increased concentration, pressure, or temperature would increase the number of collisions per unit time.

The overall kinetic energy of the system is determined by its temperature, not by the presence of a catalyst.

This result comes from incorrectly multiplying the theoretical volume by the efficiency ($15.7 \text{ L} \times 0.25$), rather than dividing by it to account for the extra fuel needed due to energy loss.

This value is obtained by incorrectly multiplying the theoretical volume by $0.75$ (or $100\% - 25\%$), which is an incorrect application of the efficiency percentage.

This is the volume of octane required if the engine were $100\%$ efficient ($528 \text{ MJ} / 33.67 \text{ MJ/L}$). It fails to account for the $25.0\%$ efficiency of the engine, which requires more fuel to be burned.

Decreasing the volume of the vessel increases the concentration of the reactant gases, which increases the rate of the forward reaction.

Heating the reaction vessel increases the kinetic energy of the molecules, which increases the rate of both the forward and reverse reactions.

Adding a catalyst lowers the activation energy of the reaction, which increases the rate of both the forward and reverse reactions.

At equilibrium, the forward and reverse reactions continue to occur at equal rates, meaning the rates are not zero. This is known as dynamic equilibrium.

While the concentrations of reactants and products remain constant at equilibrium, they do not have to be equal to each other.

Macroscopic properties, including temperature, remain constant when a system is at equilibrium.

Cellular respiration releases energy, making it an exothermic reaction, not an endothermic one.

The products of cellular respiration are carbon dioxide ($\text{CO}_2$) and water ($\text{H}_2\text{O}$), not solid carbon.

In cellular respiration, glucose is oxidized (loses electrons) while oxygen is reduced (accepts electrons).

The shape of the energy profile diagram changes because the peak, which represents the activation energy, is lowered when a catalyst is added.

The x-axis of an energy profile diagram represents reaction progress, not time. A catalyst lowers the peak vertically rather than shifting it horizontally.

A catalyst lowers the activation energy of both the forward and reverse reactions by the same absolute amount, not the same proportion, since their initial activation energies are different.

The mass of $\text{SO}_2$ relative to the remaining $\text{SO}_3$ depends on the position of equilibrium (the extent of the reaction), which changes with temperature.

According to the stoichiometry of the reaction, two moles of $\text{SO}_2$ are produced for every one mole of $\text{O}_2$, so their molar amounts will always be in a 2:1 ratio, not equal.

While photosynthesis is a redox reaction, it requires an input of light energy, making it endothermic rather than exothermic.

Photosynthesis requires energy input (endothermic) rather than releasing it (exothermic), and its overall mechanism is a redox process, not a condensation reaction.

Although photosynthesis is endothermic, the overall process is a redox reaction where $CO_2$ is reduced and $H_2O$ is oxidized, not a condensation reaction.

The Haber process requires high pressure, not low pressure, because high pressure shifts the equilibrium toward the product side (fewer moles of gas) and increases the reaction rate.

While the Haber process uses high pressure, using a purely "high" temperature for either process would drastically reduce the yield of the exothermic reactions. Instead, a compromise temperature is required.

In the Contact process step $2 SO_2(g) + O_2(g) \leftrightharpoons 2 SO_3(g)$, there are 3 moles of gas on the left and 2 on the right. Increasing pressure shifts the equilibrium to the right, thereby increasing (not reducing) the yield of $SO_3(g)$.