The decomposition of dimethyl ether leads to the formation of $CH_4$, $H_2$ and $CO$ and the reaction rate is given by Rate $= k[CH_3OCH_3]^{3/2}$ The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate $= k\,(p_{CH_3OCH_3})^{3/2}$ If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?

> [!TLDR] > The rate of a reaction is always expressed as change in concentration or pressure per unit time. For the given reaction, the rate has units of $\text{bar min}^{-1}$, and the rate constant $k$ for a $3/2$ order reaction has units of $\text{bar}^{-1/2} \text{min}^{-1}$. The key here is to understand what "rate" means in the context of a gas-phase reaction being followed by pressure change. The problem gives you two equivalent forms of the rate law — one in terms of concentration and one in terms of partial pressure. Since the question asks for units when pressure is in bar and time in minutes, we work with the pressure-based expression. Let’s break it down step by step. 1. **Identify what "rate" means in this context.** In chemical kinetics, the rate of a reaction is defined as the change in concentration (or partial pressure, for gases) of a reactant or product per unit time. Here, the reaction is followed by measuring the increase in total pressure in a closed vessel. The rate is given as: $$\text{Rate} = k\,(p_{CH_3OCH_3})^{3/2}$$ This means the rate itself has units of pressure per time — specifically, bar per minute. 2. **Determine the units of rate.** Since pressure is measured in bar and time in minutes, the rate must have units of: $$\text{Units of rate} = \text{bar min}^{-1}$$ This is straightforward: rate is a change in pressure over time. 3. **Set up the dimensional equation for the rate constant.** From the rate law: $$\text{Rate} = k \times (p)^{3/2}$$ Substitute the units we know: $$\text{bar min}^{-1} = \text{units of } k \times (\text{bar})^{3/2}$$ To isolate the units of $k$, divide both sides by $(\text{bar})^{3/2}$: $$\text{units of } k = \frac{\text{bar min}^{-1}}{\text{bar}^{3/2}} = \text{bar}^{1 - 3/2} \text{min}^{-1} = \text{bar}^{-1/2} \text{min}^{-1}$$ 4. **Check the logic with a familiar example.** …

The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by

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Concept understanding — Zero Order Kinetics

Zero Order Kinetics: The Drug That Doesn't Care How Much You Give It

Imagine you're filling a bathtub. You turn the tap to a fixed flow rate — say, 5 litres per minute. The amount of water in the tub increases by exactly 5 litres every minute, regardless of whether the tub is empty or already half full. That's the core intuition behind zero order kinetics: a constant amount disappears per unit time, no matter how much is left.

Now contrast this with what you probably expect. Most processes in nature follow first order kinetics: the rate depends on how much is present. If you have 100 molecules, 10 might react per second; if you have 10 molecules, only 1 reacts per second. The fraction lost is constant, but the amount lost per second shrinks as the quantity shrinks. Zero order is the opposite — the amount lost per second is fixed, so the fraction lost actually increases as the quantity drops.

The Precise Statement

-\frac{d[A]}{dt} = k_0

Where $[A]$ is the concentration of the substance (or amount, depending on context), $t$ is time, and $k_0$ is the zero order rate constant with units of concentration per time (e.g., mg/L per hour, or simply mg/hour if we're talking about total amount).

The negative sign indicates the substance is being removed. The key point: the rate does not depend on $[A]$ . It's a flat, constant rate.

The Integrated Form and Half-Life

If you integrate the differential equation, you get a straight line:

[A]_t = [A]_0 - k_0 t

This is the equation of a line with slope $-k_0$ and intercept $[A]_0$ . Plot concentration vs. time, and you get a straight line sloping downward until it hits zero.

The half-life — the time for the concentration to fall to half its initial value — is:

t_{1/2} = \frac{[A]_0}{2k_0}

Notice something crucial: the half-life depends on the initial concentration. Double the starting amount, and the half-life doubles. This is completely different from first order kinetics, where half-life is constant regardless of starting concentration.

Watch out

A common mistake: students assume half-life is always constant. For zero order, it is not. The half-life changes with the starting amount. If you start with 100 mg, half-life might be 5 hours; start with 200 mg, half-life becomes 10 hours.

Where Does Zero Order Kinetics Actually Happen?

In pharmacology, zero order kinetics is most famously seen with ethanol (alcohol) and aspirin at high doses. The reason is saturation of enzymes.

Your body metabolises alcohol using an enzyme called alcohol dehydrogenase. At low alcohol levels, the enzyme works efficiently and the rate depends on how much alcohol is present (first order). But at higher concentrations — say, after a few drinks — the enzyme becomes saturated. It's working at maximum speed, like a factory running at full capacity. Adding more raw material (alcohol) doesn't make it work faster. The rate becomes constant: a fixed amount of alcohol is metabolised per hour, regardless of how much is in your blood.

This is why alcohol elimination follows a straight line when you plot blood alcohol concentration vs. time. A typical person eliminates about 0.015 g/dL per hour — a fixed amount, not a fixed fraction.

Note

The same saturation principle applies to some drug transporters in the kidneys. When the transport proteins are working at maximum capacity, drug excretion becomes zero order. This is why high doses of certain drugs (like phenytoin) can lead to unexpectedly long elimination times — the system is overwhelmed.

A Quick Comparison Table

Property	Zero Order	First Order
Rate depends on	Nothing (constant)	Concentration
Rate equation	$-\frac{d[A]}{dt} = k_0$	$-\frac{d[A]}{dt} = k_1[A]$

Why this formula?

Zero Order Kinetics: Why the Formula Holds

Let's build this from the ground up — understanding the why before the what.

The Core Idea

Zero order kinetics describes a process where the rate is constant — it does not depend on the concentration of the reactant.

This is the definition, but why would that ever happen?

Why the Rate is Constant

Imagine a reaction happening on a solid surface (like a catalyst or a tablet dissolving). The reactant molecules must first adsorb onto the surface before reacting.

If the surface is saturated with reactant molecules, adding more reactant in solution doesn't help — the surface is already full.
The reaction proceeds at a fixed speed determined by how fast the surface can process the adsorbed molecules.

Key insight: The rate is limited by the surface, not by how much reactant is floating around.

Deriving the Zero Order Rate Law

Step 1: Write the rate definition

For a reaction $A \rightarrow \text{products}$ , the rate of disappearance of $A$ is:

-\frac{d[A]}{dt} = k

where $k$ is the zero order rate constant (units: concentration/time, e.g., $\text{mol L}^{-1} \text{s}^{-1}$ ).

Notice: No $[A]$ term on the right side — that's the signature of zero order.

Step 2: Separate variables and integrate

-d[A] = k \, dt

Integrate from initial time $t=0$ (concentration $[A]_0$ ) to time $t$ (concentration $[A]_t$ ):

-\int_{[A]_0}^{[A]_t} d[A] = k \int_0^t dt

-[A]_t + [A]_0 = kt

Step 3: Rearrange to the familiar form

\boxed{[A]_t = [A]_0 - kt}

This is the integrated rate law for zero order kinetics.

What This Formula Tells Us

Linear decrease: Concentration falls linearly with time (not exponentially like first order).
Slope = $-k$ : A plot of $[A]_t$ vs. $t$ gives a straight line with slope $-k$ .
Half-life depends on initial concentration:

Set $[A]_t = \frac{[A]_0}{2}$ :

\frac{[A]_0}{2} = [A]_0 - k t_{1/2}

\boxed{t_{1/2} = \frac{[A]_0}{2k}}

Critical exam point: Unlike first order (where $t_{1/2}$ is constant), zero order half-life increases with higher initial concentration.

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The rate of a reaction is always expressed as change in concentration or pressure per unit time. For the given reaction, the rate has units of $\text{bar min}^{-1}$ , and the rate constant $k$ for a $3/2$ order reaction has units of $\text{bar}^{-1/2} \text{min}^{-1}$ .

The key here is to understand what "rate" means in the context of a gas-phase reaction being followed by pressure change. The problem gives you two equivalent forms of the rate law — one in terms of concentration and one in terms of partial pressure. Since the question asks for units when pressure is in bar and time in minutes, we work with the pressure-based expression.

Let’s break it down step by step.

Identify what "rate" means in this context. In chemical kinetics, the rate of a reaction is defined as the change in concentration (or partial pressure, for gases) of a reactant or product per unit time. Here, the reaction is followed by measuring the increase in total pressure in a closed vessel. The rate is given as:

\text{Rate} = k\,(p_{CH_3OCH_3})^{3/2}

This means the rate itself has units of pressure per time — specifically, bar per minute.

Determine the units of rate. Since pressure is measured in bar and time in minutes, the rate must have units of:

\text{Units of rate} = \text{bar min}^{-1}

This is straightforward: rate is a change in pressure over time.

Set up the dimensional equation for the rate constant. From the rate law:

\text{Rate} = k \times (p)^{3/2}

Substitute the units we know:

\text{bar min}^{-1} = \text{units of } k \times (\text{bar})^{3/2}

To isolate the units of $k$ , divide both sides by $(\text{bar})^{3/2}$ :

\text{units of } k = \frac{\text{bar min}^{-1}}{\text{bar}^{3/2}} = \text{bar}^{1 - 3/2} \text{min}^{-1} = \text{bar}^{-1/2} \text{min}^{-1}

Check the logic with a familiar example. …

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✗ Common Mistake #1: Confusing rate units for gas-phase reactions

The error: Students write the rate unit as $\text{bar min}^{-1}$ but forget that the rate here is defined in terms of pressure change per time, not concentration change.

Why it happens: In solution kinetics, rate is $\text{mol L}^{-1} \text{s}^{-1}$ . Here, the reaction is followed by pressure increase, so rate is $\frac{dp}{dt}$ .

✓ How to avoid:

Always check how the rate is defined in the problem.

If rate $= -\frac{d[C]}{dt}$ → units are $\text{concentration} \times \text{time}^{-1}$
If rate $= -\frac{dp}{dt}$ → units are $\text{pressure} \times \text{time}^{-1}$

Here, pressure is in bar and time in minutes, so:

Rate units = $\text{bar min}^{-1}$

✗ Common Mistake #2: Using the wrong order to find $k$ units

The error: Students plug $n = 3/2$ into the formula

$\text{units of } k = (\text{concentration})^{1-n} \, \text{time}^{-1}$

but use $\text{mol L}^{-1}$ instead of bar.

Why it happens: The formula is memorised for concentration-based rates, but here pressure replaces concentration.

✓ How to avoid:

Use the general dimensional method:

\text{Rate} = k \, (p)^{3/2}

So:

k = \frac{\text{Rate}}{(p)^{3/2}}

Substitute units:

k = \frac{\text{bar min}^{-1}}{\text{bar}^{3/2}} = \text{bar}^{1 - 3/2} \, \text{min}^{-1} = \text{bar}^{-1/2} \, \text{min}^{-1}

Units of $k$ = $\text{bar}^{-1/2} \, \text{min}^{-1}$

✗ Common Mistake #3: Forgetting that partial pressure is used, not total pressure

The error: Students use total pressure in the rate law instead of $p_{CH_3OCH_3}$ .

Why it happens: The problem says “followed by increase in pressure” — but the rate law explicitly uses partial pressure of dimethyl ether.

✓ How to avoid:

Read carefully:

The rate is given as $k (p_{CH_3OCH_3})^{3/2}$ …

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Concept understanding — Zero Order Kinetics

Zero Order Kinetics: The Drug That Doesn't Care How Much You Give It

The Precise Statement

The Integrated Form and Half-Life

Where Does Zero Order Kinetics Actually Happen?

A Quick Comparison Table

Why this formula?

Zero Order Kinetics: Why the Formula Holds

The Core Idea

Why the Rate is Constant

Deriving the Zero Order Rate Law

Step 1: Write the rate definition

Step 2: Separate variables and integrate

Step 3: Rearrange to the familiar form

What This Formula Tells Us

Method: Dimensional Analysis for Rate and Rate Constant Units

Step 1: Write the definition of rate in terms of pressure

Step 2: Write the given rate law

Step 3: Solve for the unit of $k$

✗ Common Mistake #1: Confusing rate units for gas-phase reactions

✗ Common Mistake #2: Using the wrong order to find $k$ units

✗ Common Mistake #3: Forgetting that partial pressure is used, not total pressure

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