The main idea is that Vmax is reached when every enzyme molecule is actively turning substrate into product as fast as possible. The rate of product formation in the catalytic step is determined by k2, the rate constant for ES turning into E and P. Each enzyme molecule contributes k2 to the rate, so the total rate is k2 times how many enzyme molecules are present, i.e., k2 [ET]. At saturating substrate, all enzyme is in the ES form, so [ES] equals [ET], and Vmax = k2 [ET]. This shows Vmax depends on how much enzyme there is and the catalytic speed, but not on substrate concentration.

For a first-order process, the half-life is t1/2 = ln 2 / k, so it depends on the rate constant k but not on the initial amount. Temperature affects k through the Arrhenius relationship: higher temperature raises k. Therefore, as temperature increases, the half-life decreases. An example like a thermal decomposition that follows first order fits this perfectly: heating the sample speeds up the reaction (larger k), so you reach half of the reactant remaining sooner (a smaller t1/2). The other descriptions don’t align with this behavior. A second-order process has t1/2 that depends on the initial concentration and, although k increases with temperature, the stated trend of t1/2 increasing with temperature isn’t generally correct. A zero-order process would have t1/2 that changes with k and initial concentration, not a constant value, so saying it’s constant with temperature is inaccurate. And a first-order process cannot have t1/2 increasing with temperature because t1/2 is inversely related to k, which grows with temperature.

In a reversible reaction, equilibrium occurs when the forward and reverse rates balance each other. The forward rate is k_forward times the concentration of A, and the reverse rate is k_reverse times the concentration of B. At equilibrium, k_forward[A] = k_reverse[B]. Rearranging gives [B]/[A] = k_forward/k_reverse. The equilibrium constant for this 1:1 system is defined as K = [B]/[A], so K equals the ratio of the forward to reverse rate constants, and at equilibrium [B]/[A] = K.

Temperature changes affect the rate constant through the Arrhenius form k = A e^(-Ea/RT). Here Ea is the activation energy and A is the frequency factor. As temperature increases, the ratio Ea/RT decreases, making the exponent less negative. A less negative exponent means the exponential term grows, so k increases. Physically, more molecules have enough energy to overcome the activation barrier, leading to more successful collisions per unit time. Since the exponential term is always positive, k cannot become negative, and increasing temperature does not make k stay the same or decrease.

Reaction rate is the speed at which a reaction proceeds, i.e., how fast concentrations change with time. It’s defined as the change in concentration of a reactant (negative sign) or a product (positive sign) per unit time. For a simple reaction A → products, the rate can be written as rate = -d[A]/dt = d[Product]/dt. The usual units are molarity per second (M/s). Rate can be described as an average rate over a time interval or as an instantaneous rate at a particular moment. The other ideas describe different things: the product of concentrations isn’t the rate itself, the equilibrium constant tells how far the reaction goes at equilibrium, and the energy change per mole is a thermodynamic property, not the speed of the reaction.