In the following reaction; xA - yB
log_10[-d[A]/dt] = log_10[d[B]/dt] + 0.3010
A and B respectively

Q: Why did we suddenly replace $0.3010$ with $\log_{10} 2$? How was I supposed to know that?

In competitive exams like JEE, you are expected to memorize the values of common logarithms, specifically $\log_{10} 2 \approx 0.3010$ and $\log_{10} 3 \approx 0.4771$. Recognizing these numbers allows you to convert constants back into logarithmic form, which is often the key to simplifying complex equations using log properties.

Q: What exactly does the negative sign in $-\frac{d[A]}{dt}$ mean? Can a rate be negative?

The rate of a reaction is always a positive quantity. The term $\frac{d[A]}{dt}$ represents the change in concentration of reactant A over time. Since A is being consumed, its concentration is decreasing, making $\frac{d[A]}{dt}$ a negative value. We add the negative sign in front ($-\frac{d[A]}{dt}$) to mathematically ensure that the overall expression for the rate of disappearance remains a positive number.

Q: I don't understand how we went from $-\frac{1}{2} \frac{d[A]}{dt} = \frac{d[B]}{dt}$ to the reaction $2A \rightarrow B$. Can you explain the connection?

The standard way to write the rate of a reaction $aA \rightarrow bB$ is to divide the rate of change of each substance by its stoichiometric coefficient: $\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt}$. By comparing our derived equation $-\frac{1}{2}\frac{d[A]}{dt} = \frac{1}{1}\frac{d[B]}{dt}$ to this standard form, we can directly read off the coefficients: $a=2$ and $b=1$. This means 2 moles of A react to form 1 mole of B.

Q: Why is option (d) $N_2O_4 \rightarrow 2NO_2$ incorrect? It has a 2 and a 1 in it.

While option (d) does involve the numbers 1 and 2, the order is reversed. The reaction $N_2O_4 \rightarrow 2NO_2$ means 1 mole of reactant forms 2 moles of product (a 1:2 ratio). Our derived equation requires 2 moles of reactant to form 1 mole of product (a 2:1 ratio), which is why $2C_2H_4 \rightarrow C_4H_8$ is the correct answer.

Q: What if the base of the logarithm wasn't 10? Would the process be different?

The mathematical process would be exactly the same as long as all logarithms in the equation share the same base. The properties of logarithms (like the product rule and taking the antilog) apply universally regardless of the base. However, the specific constant value would change; for example, $\ln 2$ (base $e$) is approximately $0.693$.

Question

In the following reaction; $xA \rightarrow yB$
$\log_{10} \left[ -\frac{d[A]}{dt} \right] = \log_{10} \left[ \frac{d[B]}{dt} \right] + 0.3010$
$A$ and $B$ respectively can be

Options:

(A) $n$-butane and $iso$-butane

(B) $C_2H_2$ and $C_6H_6$

(C) $C_2H_4$ and $C_4H_8$

(D) $N_2O_4$ and $NO_2$

Sigma Percentile · Accepted Answer

### The Mystery of the Logarithmic Rate Equation

Imagine you are a detective, and you've just been handed a cryptic note. The note doesn't give you the identity of the suspects directly; instead, it gives you a complex mathematical relationship between their movements. This is exactly what we face in this intriguing chemical kinetics problem. We are given a logarithmic equation relating the rate of disappearance of a mysterious reactant $A$ to the rate of appearance of an unknown product $B$.

The equation looks like this:

$$\log_{10} \left[ -\frac{d[A]}{dt} ight] = \log_{10} \left[ \frac{d[B]}{dt} ight] + 0.3010$$

At first glance, this might seem terrifying. Logarithms mixed with calculus? But don't get intimidated. Let's take a breath and break this down step-by-step. Our ultimate goal is to strip away the mathematical disguise and reveal the simple, underlying stoichiometry of the reaction, which will then allow us to identify the actual chemicals involved.

### The Magic Number: 0.3010

In the world of competitive exams like JEE and NEET, certain numbers are like old friends; when you see them, you should immediately recognize them. Look closely at the constant term on the right side of the equation: **$0.3010$**.

Why is this number so special? Because it is the standard, widely used approximate value for the base-10 logarithm of 2.

$$\log_{10} 2 \approx 0.3010$$

This is a huge hint! By recognizing this, we can substitute $\log_{10} 2$ back into our equation. This transforms a messy equation with a random decimal into a clean equation where every single term is a logarithm.

$$\log_{10} \left[ -\frac{d[A]}{dt} ight] = \log_{10} \left[ \frac{d[B]}{dt} ight] + \log_{10} 2$$

This is a critical strategic move. It sets the stage for us to use the powerful properties of logarithms to simplify the expression further.

### Unlocking the Equation with Log Properties

Now that all our terms are logarithms with the same base (base 10), we can use the fundamental product rule of logarithms. Remember the rule?

$$\log m + \log n = \log(m 	imes n)$$

We can apply this rule to the right side of our equation to elegantly combine the two separate logarithmic terms into a single, unified term.

$$\log_{10} \left[ -\frac{d[A]}{dt} ight] = \log_{10} \left( 2 	imes \left[ \frac{d[B]}{dt} ight] ight)$$

Look at how much cleaner that is! Because we now have a single logarithm on both sides of the equals sign, and they share the exact same base, we can perform the mathematical equivalent of a magic trick: we take the antilog on both sides. This completely strips away the logarithms, leaving us with a beautifully simple, linear relationship between the rates.

$$-\frac{d[A]}{dt} = 2 \frac{d[B]}{dt}$$

### The Big Reveal: Finding the Stoichiometry

We are almost there. To truly understand what this equation is telling us about the reaction, we need to rearrange it to match the standard textbook format for expressing the rate of a chemical reaction. We achieve this by simply dividing both sides by 2.

$$-\frac{1}{2} \frac{d[A]}{dt} = \frac{d[B]}{dt}$$

Now, let's bring in the general theory. For a generic chemical reaction where $x$ moles of $A$ react to form $y$ moles of $B$ ($xA ightarrow yB$), the rate of the reaction is defined as:

$$	ext{Rate} = -\frac{1}{x} \frac{d[A]}{dt} = \frac{1}{y} \frac{d[B]}{dt}$$

Notice the beautiful symmetry here. The coefficients in the denominator of the rate expression correspond directly to the stoichiometric coefficients in the balanced chemical equation. By comparing our derived equation with this general formula, the truth is finally revealed:

**$x = 2$** and **$y = 1$**

This means that for every two moles of reactant $A$ that disappear, exactly one mole of product $B$ is formed. Our mystery reaction has a simple 2:1 stoichiometry:

**$2A ightarrow B$**

### Analyzing the Suspects

With the stoichiometry unmasked as $2A ightarrow B$, we can now evaluate our lineup of suspects (the given options) to find the perfect match.

*   **(a) $n$-butane $ightarrow iso$-butane:** This is a simple isomerization. One molecule of $n$-butane rearranges to form one molecule of $iso$-butane. The stoichiometry is 1:1. This is not our match.
*   **(b) $3C_2H_2 ightarrow C_6H_6$:** This is the famous trimerization of acetylene to form benzene. Three molecules of acetylene combine to form one molecule of benzene. The stoichiometry is 3:1. Incorrect.
*   **(c) $2C_2H_4 ightarrow C_4H_8$:** This is the dimerization of ethylene to form butene. Exactly two molecules of ethylene combine to form one molecule of butene. The stoichiometry is 2:1. **This is a perfect match!**
*   **(d) $N_2O_4 ightarrow 2NO_2$:** This is the decomposition of dinitrogen tetroxide. One molecule breaks apart to form two molecules of nitrogen dioxide. The stoichiometry is 1:2. Incorrect.

Therefore, the only reaction that fits the mathematical profile we uncovered is the dimerization of ethylene. The correct option is indeed **(c)**.

This problem is a fantastic example of how abstract mathematical relationships in kinetics are deeply tied to the physical reality of how molecules interact and combine.

Animated Solution for Chemistry - Chemical Kinetics: In the following reaction; $x A \to y B$ $lo g_{10} [- \frac{d [ A ]}{d t}] = lo g_{10} [\frac{d [ B ]}{d t}] + 0.3010$ $A$ and $B$ respectively can be

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Visualized Solution

$Analyzing the Given Equation$

$Identifying the Constant$

$Substituting the Logarithmic Value$

$Applying Logarithm Properties$

$Removing Logarithms$

$Rearranging to Standard Rate Form$

$Determining the Stoichiometry$

$Evaluating the Options$

The Sigma Insight: Rate of Chemical Reaction

The Mystery of the Logarithmic Rate Equation

The Magic Number: 0.3010

Unlocking the Equation with Log Properties

The Big Reveal

Analyzing the Suspects

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Animated Solution for Chemistry - Chemical Kinetics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the following reaction; xA→yB log10​[−dtd[A]​]=log10​[dtd[B]​]+0.3010 A and B respectively can be

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Visualized Solution

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearranging to Standard Rate Form

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

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