For a first-order reaction, the time to reach 25% of the initial concentration is t = (ln(1/0.25))/k = (ln(4))/0.5 = 2.772/0.5 = 5.544 seconds.

An increase in activation energy decreases the rate constant according to the Arrhenius equation, k = Ae^(-Ea/RT), where an increase in Ea results in a smaller k.

According to the Arrhenius equation, an increase in temperature results in an increase in the rate constant, k, as it provides more energy to overcome the activation energy barrier.

According to the Arrhenius equation, an increase in temperature generally increases the rate constant due to the exponential dependence on temperature.

Using the Arrhenius equation, k2 = k1 * e^[-Ea/R(1/T2 - 1/T1)]. Substituting values gives k2 ≈ 0.4 s^-1.

Heat absorbed = ΔH * n = -120 kJ * 0.25 = -30 kJ.

W = -nFE° = -3 * 96485 C/mol * 0.45 V = -130.9 kJ.

The stronger oxidizing agent is the one with the higher reduction potential. Here, A/B has a higher E° than C/D.

In a zero-order reaction, the rate is constant and does not depend on the concentration of the reactants.

In a zero-order reaction, the rate is constant, leading to a linear decrease in concentration over time.

For a zero-order reaction, [A] = [A]0 - kt. Thus, 0.2 M = 0.5 M - (0.1 M/s)t, leading to t = (0.5 - 0.2) / 0.1 = 3 s.

In a zero-order reaction, the rate is constant, leading to a linear decrease in concentration over time.

The first law of thermodynamics states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W).

Removing O2 will shift the equilibrium to the left to produce more O2, in accordance with Le Chatelier's Principle.

If the reaction is exothermic, increasing the temperature will shift the equilibrium to the left, favoring the formation of reactants (NO2).

Removing O2 decreases its concentration, causing the equilibrium to shift to the left to produce more O2, in accordance with Le Chatelier's Principle.

Removing SO3 decreases its concentration, causing the equilibrium to shift to the right to produce more SO3 and restore balance.

Increasing the volume decreases the pressure, which shifts the equilibrium to the left, favoring the side with more moles of gas (3 moles of reactants).

Increasing pressure favors the side with fewer moles of gas. If A and B have more moles than C, the equilibrium will shift to the left.

Decreasing the temperature of an endothermic reaction shifts the equilibrium to the right, favoring the formation of products.

Increasing the concentration of H2 will shift the equilibrium to the right, favoring the production of CH3OH.

Using the expression Kc = [NO2]^2 / ([NO]^2[O2]), we find that at equilibrium [NO2] = 0.4 M.

Removing SO3 decreases its concentration, causing the equilibrium to shift to the right to produce more SO3, according to Le Chatelier's Principle.

Increasing the pressure will shift the equilibrium to the right, favoring the formation of SO3, as it has fewer moles of gas.

Increasing the pressure will shift the equilibrium to the right, favoring the production of SO3, as it has fewer moles of gas (3 moles to 2 moles).

Decreasing the temperature of an exothermic reaction will shift the equilibrium to the right, favoring the formation of SO3.

If the reaction is exothermic, decreasing the temperature will shift the equilibrium to the right, favoring NH3 production.

Increasing the temperature of an exothermic reaction shifts the equilibrium to the left, favoring the reactants.

A negative ΔH indicates that the reaction releases heat, thus it is exothermic.

At equilibrium, ΔG = 0; ΔG = ΔH - TΔS = 0; T = ΔH/ΔS = 150,000 J / 200 J/K = 750 K.