The reaction 2A + B2 - 2AB is an elementary reaction. For a certain quantity of reactants, if the v

Q: Why did we assume the order of reaction is 2 for A and 1 for B2?

The magic word in the question is 'elementary'. For elementary reactions, the reaction happens in a single step, meaning the molecularity equals the order. Therefore, the stoichiometric coefficients directly become the powers in the rate law.

Q: What if the question didn't say 'elementary'?

We would be stuck! For complex reactions, the stoichiometric coefficients do not necessarily match the order of the reaction. We would need experimental data (like a table of initial rates) to determine the true rate law.

Q: Why does reducing the volume increase the concentration?

Concentration is defined as moles per unit volume ($C = n/V$). Since volume is in the denominator, decreasing the volume while keeping the moles constant mathematically increases the concentration. Physically, the particles are forced into a smaller space, making them more crowded.

Q: Does the rate constant 'k' change when we change the volume?

No! The rate constant $k$ is very stubborn. It only changes if you change the temperature of the system or if you add a catalyst. Changing volume or pressure does not affect $k$.

Q: What happens to the number of moles during this compression?

The question explicitly states 'For a certain quantity of reactants'. This implies that the number of moles ($n_A$ and $n_B$) remains constant during the instant we are comparing the rates.

Question

The reaction $2A + B_2 \rightarrow 2AB$ is an elementary reaction. For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ........... . (Round off to the nearest integer).

Sigma Percentile · Accepted Answer

### The Magic of Elementary Reactions: Volume and Rate

Imagine you are holding a syringe filled with a mixture of two reacting gases, $A$ and $B_2$. The particles are buzzing around, colliding, and occasionally reacting to form the product $2AB$. Now, what happens if you suddenly push the plunger down, compressing the gases into a much smaller space?

Intuitively, you know the particles will be more crowded. They will bump into each other much more frequently, and the reaction will speed up. But in chemistry, we don't just want to know *if* it speeds up; we want to know *exactly by how much*. This is where the magic of the rate law comes into play.

### The Elementary Advantage

The question gives us a massive hint right at the beginning: **"The reaction is an elementary reaction."**

Why is this so important? In chemical kinetics, an elementary reaction is a reaction that occurs in a single, simple step. There are no hidden intermediates or complex mechanisms. Because it's a one-step dance, the stoichiometry of the balanced equation directly dictates the order of the reaction.

For the reaction $2A + B_2 \rightarrow 2AB$, we can confidently write the rate law as:

$$r = k[A]^2[B_2]$$

Notice how the coefficient $2$ for reactant $A$ becomes the power of $2$, and the coefficient $1$ for $B_2$ becomes the power of $1$. If this were a complex reaction, we would be completely stuck without experimental data!

### The Math of Compression

Let's define our initial state. Concentration is simply the number of moles ($n$) divided by the volume ($V$). So, our initial rate $r_1$ is:

$$r_1 = k \left(\frac{n_A}{V}\right)^2 \left(\frac{n_B}{V}\right)$$

Now, we compress the vessel, reducing its volume by a factor of 3. The new volume is $V' = \frac{V}{3}$.

Because the volume is in the denominator, dividing the volume by 3 is mathematically identical to multiplying the concentration by 3. The particles are three times as crowded!

Our new concentrations are:
$$[A]' = \frac{n_A}{V/3} = 3[A]$$
$$[B_2]' = \frac{n_B}{V/3} = 3[B_2]$$

### The Final Calculation

Let's substitute these new, tripled concentrations back into our master rate law to find the new rate, $r_2$:

$$r_2 = k (3[A])^2 (3[B_2])$$

When we expand this, we must be careful to square the $3$ inside the $A$ term:

$$r_2 = k \cdot 9[A]^2 \cdot 3[B_2]$$

$$r_2 = 27 \cdot k[A]^2[B_2]$$

Since $k[A]^2[B_2]$ is our initial rate $r_1$, we can clearly see that:

$$r_2 = 27 r_1$$

By simply reducing the volume by a factor of 3, the reaction rate skyrocketed by a factor of **27**. This dramatic increase highlights the extreme sensitivity of higher-order reactions to changes in concentration and volume. Always keep an eye out for that word "elementary"—it's the key that unlocks the whole problem!

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Reaction Vessel}

\text{Rate Law for Elementary Reaction}

\text{Initial Rate Expression}

\text{Reducing the Volume}

\text{Final Rate Expression}

\text{Calculating the Factor}

\text{Key Takeaway}

The Sigma Insight: Rate of Chemical Reaction

The Magic of Elementary Reactions

The Elementary Advantage

The Math of Compression

The Final Calculation

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