Q.In general, what happens to the rate of reaction as the reaction progresses?

The assertion is false because order is defined only for simple rate laws (elementary steps or overall reactions with a simple rate expression), not for all complex reactions. The reason is true: order is experimental. So the correct option is (D).

For a third-order reaction, Rate ∝ [X]^3, so tripling [X] multiplies the rate by 3^3 = 27.

Rate law: Rate = k[X]^3 (third order in X).

Original rate: Rate1 = k[X]^3 …

Rate depends on active mass (concentration), per the law of mass action.

The law of mass action states that the rate of a chemical reaction is proportional to the product of the ACTIVE MASSES (molar concentrations) of the reacting substances, each raised to a power. Atomic, equivalent and molecular masses are fixed properties of a substance and do not govern how fast a reaction …

Second order kinetics means Rate = k[R]^2; scaling the concentration by a factor scales the rate by that factor squared.

Rate = k[R]^2 (second order in R)

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The rate constant $k$ is an intrinsic property of a reaction at a given temperature — it does NOT depend on the concentrations of reactants; only the rate itself changes when concentration changes.

For Rate $= k[A][B]$, doubling $[B]$ while keeping $[A]$ constant doubles the RATE of the reaction (since rate $\propto [B]$), but $k$ itself is unaffected — $k$ is a constant determined only by the nature of the reaction, the temperature, and the …

As reactants are used up over time, their concentration drops, and since rate depends on concentration, the rate falls too.

For most reactions, the rate law has the form $Rate = k[\text{reactant}]^n$ (n > 0), i.e. the rate is directly related to the concentration of the reactant(s).

As a reaction proceeds, reactant molecules are continuously converted to products, so the concentration of the reactants keeps decreasing with time. Since the rate depends on this concentration, the rate of reaction generally decreases as the reaction progresses, being fastest at the very start (when reactant concentration is highest) and slowing down as the reaction approaches completion.

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Rate depends on [B]^2, so tripling [B] increases the rate by a factor of 9.

Given Rate = K[A][B]^2 and [A] kept constant, only the [B]^2 term changes. If [B] is trebled (multiplied by 3):

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Substituting the doubled concentrations into the rate law Rate = k[A]^1[B]^2 shows the rate scales by a factor of 8.

Given: order in A = 1, order in B = 2, so Rate = k[A][B]^2.

If [A] -> 2[A] and [B] -> 2[B]: …

The rate law states that the rate of a reaction is proportional to the concentration(s) of the reactant(s) raised to some power (the order), not necessarily equal to the stoichiometric coefficient.

For a general reaction A -> products, the experimentally determined rate law is written as:

Rate = k[A]^n

where k is the rate constant and n is the order of the reaction with respect to A (found experimentally, not simply read off the balanced equation). This shows the rate of reaction is a function of (depends on / is proportional to) the concentration of the reactant — as concentration increases, the frequency of effective molecular collisions increases, so the rat …

The rate law is the experimentally determined equation expressing reaction rate as proportional to reactant concentrations raised to their (experimental) orders.

For a general reaction, $aA + bB \rightarrow \text{Products}$, the rate law (or rate equation) expresses how the rate of reaction depends on the concentration of each reactant:

$\text{Rate} = k[A]^x[B]^y$ …

The rate law connects reaction rate to reactant concentrations via experimentally-found exponents (orders).

For a reaction, the rate law (or rate equation) is the mathematical expression relating the instantaneous rate of the reaction to the molar concentrations of the reactants, each raised to a power (the order with respect to that reactant), determined experimentally (not necessarily from the stoichiometric coefficients):

$Rate = k[A]^x[B]^y$ …