NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation, 2N

Q: Why is there a negative sign in the rate of disappearance formula?

The concentration of reactants decreases over time, making the change in concentration ($\Delta[Reactant]$) a negative number. We add a negative sign in front of the formula to mathematically cancel it out, ensuring the overall rate is reported as a positive value.

Q: Why do we divide the individual rates by the stoichiometric coefficients?

We do this to define a single, unique 'Rate of Reaction' for the entire chemical equation. Because molecules appear and disappear at different speeds based on their molar ratios, dividing by the coefficients normalizes these speeds so they are all equal to one another.

Q: Can I use the initial and final concentrations of $NO_2$ directly to find its rate?

If the problem provided the initial and final concentrations of $NO_2$, yes, you absolutely could! However, in this specific problem, we are only given data for $N_2O_5$, which is why we must use stoichiometry to relate the two.

Q: What if the time was given in seconds instead of minutes?

You would calculate the rate in $\text{mol L}^{-1} \text{ s}^{-1}$. Always check the units of the multiple-choice options to see if a time conversion (like multiplying or dividing by 60) is necessary before finishing your calculation.

Q: Is the rate of formation of a product always greater than the rate of disappearance of a reactant?

Not necessarily! It depends entirely on the stoichiometric coefficients. Here, 4 moles of $NO_2$ are formed for every 2 moles of $N_2O_5$ consumed, so its rate is higher. But if we looked at $O_2$, its coefficient is 1, meaning it forms at half the speed $N_2O_5$ disappears!

Question

$	ext{NO}_2$ required for a reaction is produced by the decomposition of $	ext{N}_2	ext{O}_5$ in $	ext{CCl}_4$ as per the equation,
$2	ext{N}_2	ext{O}_5(g) \longrightarrow 4	ext{NO}_2(g) + 	ext{O}_2(g)$
The initial concentration of $	ext{N}_2	ext{O}_5$ is $3.00 	ext{ mol L}^{-1}$ and it is $2.75 	ext{ mol L}^{-1}$ after 30 minutes. The rate of formation of $	ext{NO}_2$ is

Options:

(A) $4.167 	imes 10^{-3} 	ext{ mol L}^{-1} 	ext{ min}^{-1}$

(B) $1.667 	imes 10^{-2} 	ext{ mol L}^{-1} 	ext{ min}^{-1}$

(C) $8.333 	imes 10^{-3} 	ext{ mol L}^{-1} 	ext{ min}^{-1}$

(D) $2.083 	imes 10^{-3} 	ext{ mol L}^{-1} 	ext{ min}^{-1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a sealed container where a chemical transformation is taking place. We are observing the decomposition of Dinitrogen pentoxide ($N_2O_5$) into Nitrogen dioxide ($NO_2$) and Oxygen gas ($O_2$).

The balanced chemical equation for this process is:

$$2N_2O_5(g) \longrightarrow 4NO_2(g) + O_2(g)$$

We are given a snapshot of this reaction over a specific time interval. Initially, the concentration of $N_2O_5$ is $3.00 	ext{ mol L}^{-1}$. After a span of $30 	ext{ minutes}$, the concentration drops to $2.75 	ext{ mol L}^{-1}$. Our mission is to determine how fast the product, $NO_2$, is being formed during this exact time window.

### The Rate of Disappearance

Before we can find out how fast the product is appearing, we must first calculate how fast our reactant is disappearing. The **rate of disappearance** of a reactant is defined as the negative change in its concentration divided by the time interval.

Why the negative sign? Because the final concentration is lower than the initial concentration, the change ($\Delta$) is inherently negative. The extra negative sign ensures our rate is a positive, meaningful physical quantity.

$$	ext{Rate of disappearance of } N_2O_5 = -\frac{\Delta[N_2O_5]}{\Delta t}$$

Let's plug in our raw numbers:

$$-\frac{\Delta[N_2O_5]}{\Delta t} = -\frac{(2.75 - 3.00)}{30}$$

$$-\frac{\Delta[N_2O_5]}{\Delta t} = \frac{0.25}{30} 	ext{ mol L}^{-1} 	ext{ min}^{-1}$$

### The Master Equation of Kinetics

Now, how do we bridge the gap between the reactant disappearing and the product appearing? This is where the stoichiometry of the balanced equation becomes our most powerful tool.

Notice that for every **2 moles** of $N_2O_5$ that break apart, **4 moles** of $NO_2$ are created. This means $NO_2$ is being formed twice as fast as $N_2O_5$ is being destroyed! To create a universal "Rate of Reaction" that holds true no matter which chemical species we look at, we divide the individual rates by their respective stoichiometric coefficients.

$$	ext{Rate of Reaction} = -\frac{1}{2}\frac{\Delta[N_2O_5]}{\Delta t} = +\frac{1}{4}\frac{\Delta[NO_2]}{\Delta t}$$

### Final Calculation

We want to isolate the rate of formation of $NO_2$, which is represented by the term $+\frac{\Delta[NO_2]}{\Delta t}$. By rearranging our master equation, we get:

$$\frac{\Delta[NO_2]}{\Delta t} = -\frac{4}{2}\frac{\Delta[N_2O_5]}{\Delta t}$$

$$\frac{\Delta[NO_2]}{\Delta t} = 2 	imes \left(-\frac{\Delta[N_2O_5]}{\Delta t}ight)$$

Now, we simply substitute the rate of disappearance we calculated earlier:

$$\frac{\Delta[NO_2]}{\Delta t} = 2 	imes \left(\frac{0.25}{30}ight)$$

$$\frac{\Delta[NO_2]}{\Delta t} = \frac{0.50}{30} = \frac{1}{60}$$

Converting this fraction into a decimal and then into scientific notation gives us our final, elegant answer:

**$$\frac{\Delta[NO_2]}{\Delta t} = 1.667 	imes 10^{-2} 	ext{ mol L}^{-1} 	ext{ min}^{-1}$$**

This perfectly matches option (b). The beauty of chemical kinetics lies in this exact proportionality—once you know the speed of one molecule in a reaction, the balanced equation unlocks the speeds of all the others!

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Visualized Solution

$R e a c t i o n S t o i c hi o m e t r y$

$R a t e o f D i s a pp e a r an ce o f N_{2} O_{5}$

$C a l c u l a t in g t h e R a t e$

$R a t e L a w R e l a t i o n s hi p$

$R a t e o f F or ma t i o n o f NO_{2}$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Rate of Chemical Reaction

Analyzing the Setup

The Rate of Disappearance

The Master Equation of Kinetics

Final Calculation

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate of Disappearance of N2​O5​

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate of Formation of NO2​

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

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