Q.While converting the fractional part of a decimal number to another number system, why do we write the integer part from top to bottom and not other way?

Fractional Conversion Order – From Intuition to Precision

Imagine you are watching a reaction happen. A molecule of A turns into B. You might ask: how does the rate of this reaction depend on how much A is left? That is the core question behind order — but there is a twist. Most textbook examples give you nice whole numbers: order 1, order 2. But nature is not always that tidy.

Sometimes the rate depends on concentration in a way that is not a whole number. For instance, the rate might be proportional to $[A]^{0.5}$ or $[A]^{1.3}$. That is a fractional order.

Intuition First

Think of a reaction happening on a surface — like a catalytic converter in a car. The reactant molecules stick to the surface, react, and leave. If the surface is crowded, adding more reactant does not help much; the rate becomes nearly independent of concentration (order near 0). If the surface is nearly empty, every extra molecule finds a spot easily, so the rate rises almost linearly with concentration (order near 1). In between, you get a fractional order — say 0.7 — because the relationship is not simple.

Another common source: a reaction that proceeds through multiple steps, where one step is not cleanly rate-determining. The observed rate law can end up with a fractional exponent because the mechanism involves an equilibrium that itself depends on concentration in a non-linear way.

The Precise Statement

For a reaction $aA + bB \rightarrow \text{products}$, the rate law is experimentally determined as:

Here $m$ and $n$ are the orders with respect to A and B. The overall order is $m + n$.

When $m$ or $n$ is a fraction (like $\frac{1}{2}$, $\frac{3}{2}$, 0.8), we call it a fractional order. It is not a theoretical assumption — it is what the data forces you to accept.

Why Does It Happen?

Three common reasons:

1. Surface reactions (Langmuir-Hinshelwood kinetics): The rate depends on the fraction of surface covered, which follows a curve like $\frac{K[A]}{1 + K[A]}$. At intermediate concentrations, this behaves approximately like $[A]^{0.5}$.

2. Chain reactions: In some free-radical mechanisms, the rate of product formation involves the square root of initiator concentration — giving order $\frac{1}{2}$.

3. Reversible steps before the slow step: If a fast equilibrium produces an intermediate whose concentration is proportional to $[A]^{1/2}$ (say from dissociation of a dimer), the rate law inherits that fractional exponent.

A Concrete Example

Consider the decomposition of acetaldehyde (CH$_3$CHO) into methane and carbon monoxide. Experimentally, the rate is found to be: …