For the reaction system, 2NO(g) + O2(g) - 2NO2(g) volume is suddenly reduced to half its value by i

Q: Why does halving the volume double the concentration?

Concentration is the number of moles per unit volume ($n/V$). If you divide the denominator by 2, the overall fraction is multiplied by 2. Since the number of moles hasn't changed, the molecules are just packed into half the space, doubling their density.

Q: What if the reaction was elementary? Would the answer change?

If the reaction was elementary, the order with respect to each reactant would be equal to its stoichiometric coefficient. For $2\text{NO} + \text{O}_2$, the order would be 2 for NO and 1 for O2, which is exactly what was given! So the answer would be the same.

Q: Does the rate constant $k$ change when the volume is reduced?

No, the rate constant $k$ only depends on the temperature and the presence of a catalyst. Changing the volume or pressure does not affect $k$.

Q: Why did we square the '2' for NO but not for O2?

The reaction is second order with respect to NO, meaning its concentration term is squared in the rate law. When the concentration doubles, the factor of 2 is also squared ($2^2 = 4$). For O2, it's first order, so the factor of 2 is raised to the power of 1.

Q: What would happen if the pressure was doubled instead of the volume being halved?

For an ideal gas at constant temperature, Boyle's Law states that pressure and volume are inversely proportional ($P \propto 1/V$). Doubling the pressure is exactly the same as halving the volume, so the effect on the reaction rate would be identical.

Question

For the reaction system,
$2	ext{NO}(g) + 	ext{O}_2(g) \longrightarrow 2	ext{NO}_2(g)$
volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to $	ext{O}_2$ and second order with respect to $	ext{NO}$; the rate of reaction will

Options:

(A) diminish to one-fourth of its initial value

(B) diminish to one-eighth of its initial value

(C) increase to eight times of its initial value

(D) increase to four times of its initial value

Sigma Percentile · Accepted Answer

## The Power of Compression: How Volume Affects Reaction Rates

Have you ever wondered why compressing a gas makes it react faster? It's all about bringing molecules closer together. In this problem, we explore a classic scenario where the volume of a reaction vessel is suddenly halved, and we need to find out how this impacts the rate of the reaction.

### Analyzing the Setup

We are given the reaction:
$$2	ext{NO}(g) + 	ext{O}_2(g) \longrightarrow 2	ext{NO}_2(g)$$

The problem explicitly states the order of the reaction: it is second order with respect to $	ext{NO}$ and first order with respect to $	ext{O}_2$.

Let's write down the initial rate law based on this information. The initial rate, $r_1$, can be expressed as:
$$r_1 = k[	ext{NO}]^2[	ext{O}_2]$$

Here, $k$ is the rate constant, and the square brackets denote the molar concentration of the gases.

### The Impact of Halving the Volume

Now, the volume of the vessel is suddenly reduced to half its initial value by increasing the pressure.

**What happens to the concentration?** 
Remember that molar concentration is defined as the number of moles divided by the volume ($C = \frac{n}{V}$). If the number of moles remains constant but the volume is halved ($V ightarrow V/2$), the concentration of every gas in the container must double!

So, our new concentrations become:
$$[	ext{NO}]_{	ext{new}} = 2[	ext{NO}]$$
$$[	ext{O}_2]_{	ext{new}} = 2[	ext{O}_2]$$

### Calculating the New Rate

Let's substitute these new, doubled concentrations into our rate law to find the new rate, $r_2$.

$$r_2 = k(2[	ext{NO}])^2(2[	ext{O}_2])$$

Now, we carefully expand the terms. Don't make a silly mistake with the square! The $2$ inside the squared term becomes a $4$.

$$r_2 = k \cdot 4[	ext{NO}]^2 \cdot 2[	ext{O}_2]$$

Multiplying the constants together ($4 	imes 2 = 8$), we get:

$$r_2 = 8 \cdot k[	ext{NO}]^2[	ext{O}_2]$$

### The Final Verdict

Notice that the term $k[	ext{NO}]^2[	ext{O}_2]$ is exactly our initial rate, $r_1$.

$$r_2 = 8 r_1$$

By simply halving the volume, we've forced the molecules closer together, drastically increasing their collision frequency. As a result, the reaction rate shoots up to **eight times** its original value!

This perfectly matches option (c). It's a beautiful demonstration of how physical changes like compression can exponentially accelerate chemical kinetics.

Select Answer:

Visualized Solution

$r_{1} = k [NO]^{2} [O_{2}]$

$V_{2} = \frac{V}{2}$

$r_{2} = k [NO]_{2}^{2} [O_{2}]_{2}$

$r_{2} = 8 r_{1}$

$Rate increases by 8 times$

$Elementary Reaction$

The Sigma Insight: Order and Molecularity

Analyzing the Setup

The Impact of Halving the Volume

Calculating the New Rate

The Final Verdict

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​=k[NO]2[O2​]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V2​=2V​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2​=k[NO]22​[O2​]2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2​=8r1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate increases by 8 times

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Elementary Reaction