A sample of milk splits after 60 min. at 300 K and after 40 min. at 400 K when the population of lac

Q: Why is the time taken inversely proportional to the rate constant?

For a given extent of reaction (like the population doubling), the product of the rate constant and time is a constant (e.g., $kt = \ln 2$ for a first-order reaction). Therefore, $k \propto 1/t$.

Q: Why did we use $0.4$ for $\ln(3/2)$ when the question stated $\ln(2/3) = 0.4$?

Mathematically, $\ln(2/3)$ is a negative value (approximately $-0.405$). The question provided a positive value, which is a typographical error. Since activation energy must be positive, we logically use the magnitude $0.4$ for $\ln(3/2)$.

Q: Do we need to convert the time from minutes to seconds?

No, because we are taking the ratio of the two times ($t_1/t_2$). The units cancel out, so keeping them in minutes is perfectly fine and saves calculation time.

Q: What is the significance of the phrase 'population doubles'?

It indicates that the process follows first-order kinetics, where the time taken for the population to double is the half-life (or doubling time). This confirms that the time required is strictly inversely proportional to the rate constant.

Q: Why is $e^{-3} = 4.0$ given in the question?

Sometimes, competitive exams provide extra or dummy data to test if students know exactly which values are relevant to the formula. In this case, $e^{-3} = 4.0$ is not needed for the solution.

Question

A sample of milk splits after $60	ext{ min.}$ at $300	ext{ K}$ and after $40	ext{ min.}$ at $400	ext{ K}$ when the population of lactobacillus acidophilus in it doubles. The activation energy (in $	ext{kJ/mol}$) for this process is closest to (Given, $R = 8.3	ext{ J mol}^{-1}	ext{K}^{-1}$, $\ln\left(\frac{2}{3}ight) = 0.4, e^{-3} = 4.0$)

Sigma Percentile · Accepted Answer

### The Real-World Kinetics of Spoiling Milk

Have you ever wondered why milk spoils so much faster during the hot summer months compared to the cold winter? This everyday phenomenon is a perfect demonstration of **Chemical Kinetics** in action.

In this problem, we are observing the splitting of milk, which is driven by the exponential growth of *Lactobacillus acidophilus* bacteria. We are given the time it takes for the population to double at two different temperatures: $60 	ext{ min}$ at $300 	ext{ K}$ and $40 	ext{ min}$ at $400 	ext{ K}$.

Our goal is to find the **activation energy** ($E_a$) for this biological process.

### The Master Equation: Arrhenius Law

To connect temperature, time, and activation energy, we need the legendary **Arrhenius equation**.

But first, we must establish the relationship between the time taken for the process and its rate constant ($k$). Since the population is doubling, this process follows first-order kinetics. For any given extent of a reaction, the time taken ($t$) is inversely proportional to the rate constant ($k$).

Mathematically, this means:
$$t \propto \frac{1}{k} \implies \frac{k_2}{k_1} = \frac{t_1}{t_2}$$

Now, we bring in the Arrhenius equation for two different temperatures:
$$\ln\left(\frac{k_2}{k_1}ight) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}ight)$$

Substituting our time ratio into this equation gives us our working formula:
$$\ln\left(\frac{t_1}{t_2}ight) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}ight)$$

### Substituting the Variables

Let's carefully plug in the given values. We know $t_1 = 60 	ext{ min}$ at $T_1 = 300 	ext{ K}$, and $t_2 = 40 	ext{ min}$ at $T_2 = 400 	ext{ K}$. The universal gas constant $R$ is given as $8.3 	ext{ J mol}^{-1}	ext{K}^{-1}$.

$$\ln\left(\frac{60}{40}ight) = \frac{E_a}{8.3}\left(\frac{1}{300} - \frac{1}{400}ight)$$

Now, we simplify the terms. The fraction $60/40$ reduces neatly to $3/2$.

On the right side, we find a common denominator for the temperatures:
$$\frac{1}{300} - \frac{1}{400} = \frac{400 - 300}{300 	imes 400} = \frac{100}{120000} = \frac{1}{1200}$$

This simplifies our equation to:
$$\ln\left(\frac{3}{2}ight) = \frac{E_a}{8.3 	imes 1200}$$

### The Logarithmic Trap

Here is where many students stumble. The question provides the value $\ln\left(\frac{2}{3}ight) = 0.4$.

However, mathematically, $\ln(2/3)$ is a fraction less than 1, so its natural logarithm must be negative (approximately $-0.405$). If we blindly use a negative value for $\ln(3/2)$, we would end up calculating a **negative activation energy**, which is physically impossible for this process!

We must understand the examiner's intent. The provided value is a magnitude. Therefore, we logically assume:
$$\ln\left(\frac{3}{2}ight) \approx 0.4$$

### Final Calculation

With the trap avoided, we can now solve for $E_a$:
$$0.4 = \frac{E_a}{8.3 	imes 1200}$$

Multiplying the terms across:
$$E_a = 0.4 	imes 8.3 	imes 1200 = 3984 	ext{ J mol}^{-1}$$

Finally, we convert Joules to kilojoules by dividing by $1000$:
$$E_a = 3.984 	ext{ kJ mol}^{-1}$$

Rounding to the nearest given format, we get our final answer:
**$E_a \approx 3.98 	ext{ kJ mol}^{-1}$**

This beautiful problem not only tests your grasp of the Arrhenius equation but also your presence of mind when dealing with experimental or given data!

Enter Numerical Value:

Visualized Solution

$t \propto \frac{1}{k}$

$ln (\frac{k _{2}}{k _{1}}) = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

$ln (\frac{60}{40}) = \frac{E _{a}}{8.3} (\frac{1}{300} - \frac{1}{400})$

$ln (\frac{3}{2}) = \frac{E _{a}}{8.3 \times 1200}$

$ln (\frac{3}{2}) = 0.4$

$E_{a} = 3.984 kJ mol^{- 1}$

The Sigma Insight: Theories of Chemical Reaction

The Real-World Kinetics of Spoiling Milk

The Master Equation

Substituting the Variables

The Logarithmic Trap

Final Calculation

Similar Questions

The rate of a reaction double when its temperature changes from $300 K$ to $310 K$ . Activation energy of such a reaction will be ( $R = 8.314 J K^{- 1} mol^{- 1}$ and $lo g 2 = 0.301$ )

The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from $2 7^{\circ} C$ to $4 2^{\circ} C$ . Its energy of activation in $J/mol$ is ......... . (Take $ln 5 = 1.6094$ ; $R = 8.314 J mol^{- 1} K^{- 1}$ )

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $4 0^{\circ} C$ to $3 0^{\circ} C$ . The activation energy (in $kJ mol^{- 1}$ ) of the reaction is ......... . (Take; $R = 8.314 J mol^{- 1} K^{- 1}$ , $ln 3.555 = 1.268$ )

For the reaction $A \to B$ , the rate constant $k$ (in $s^{- 1}$ ) is given by $lo g_{10} k = 20.35 - \frac{( 2.47 \times 1 0 ^{3} )}{T}$ The energy of activation in $kJ mol^{- 1}$ is ......... (Nearest integer) [Given : $R = 8.314 J K^{- 1} mol^{- 1}$ ]

The rate constant ( $k$ ) of a reaction is measured at different temperature ( $T$ ), and the data are plotted in the given figure. The activation energy of the reaction in $kJ mol^{- 1}$ is ( $R$ is gas constant)

A reactant (A) forms two products $A k_{1} B$ , Activation energy $E_{a_{1}}$ $A k_{2} C$ , Activation energy $E_{a_{2}}$ If $E_{a_{2}} = 2 E_{a_{1}}$ , then $k_{1}$ and $k_{2}$ are related as

The rate constant of a reaction increases by five times on increase in temperature from $2 7^{\circ} C$ to $5 2^{\circ} C$ . The value of activation energy in $kJ mol^{- 1}$ is ……… (Rounded off to the nearest integer) $[R = 8.314 J K^{- 1} mol^{- 1}]$

If the activation energy of a reaction is $80.9 kJ mol^{- 1}$ , the fraction of molecules at $700 K$ , having enough energy to react to form products is $e^{- x}$ . The value of $x$ is ......... (Rounded off to the nearest integer) [Use, $R = 8.31 J K^{- 1} mol^{- 1}$ ].

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t∝k1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(k1​k2​​)=REa​​(T1​1​−T2​1​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(4060​)=8.3Ea​​(3001​−4001​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(23​)=8.3×1200Ea​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ln(23​)=0.4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ea​=3.984 kJ mol−1