The first order rate constant for the decomposition of CaCO3 at 700 K is 6.36 x 10^-3 s^-1 and activ

Q: Why do we use the logarithmic form of the Arrhenius equation?

The logarithmic form allows us to easily compare rate constants at two different temperatures without dealing with complex exponential calculations directly. It turns a multiplicative relationship into a simpler additive/subtractive one.

Q: Why is it so important to convert activation energy from kJ to J?

The universal gas constant $R$ is given in units of $\text{J K}^{-1} \text{mol}^{-1}$. For the units to cancel out properly in the exponent $\frac{E_a}{RT}$, the activation energy $E_a$ must also be in Joules. Failing to convert will make your answer wrong by a factor of 1000!

Q: What does the negative slope in the Arrhenius plot signify?

The slope of the $\ln k$ vs $1/T$ plot is $-\frac{E_a}{R}$. Since activation energy ($E_a$) is always positive, the slope is always negative. This visually represents that as temperature increases ($1/T$ decreases), the rate constant $k$ increases.

Q: How do we handle the antilog calculation without a calculator?

In competitive exams like JEE, complex antilog values are usually provided in the question hints. Here, the question explicitly gave $10^{-4.79} = 1.62 \times 10^{-5}$, saving you from manual approximation.

Q: Why did the rate constant decrease when the temperature dropped?

Temperature is a measure of the average kinetic energy of molecules. When temperature drops from 700 K to 600 K, fewer molecules have enough energy to overcome the activation energy barrier, leading to fewer successful collisions and a lower rate constant.

Question

The first order rate constant for the decomposition of $	ext{CaCO}_3$ at $700	ext{ K}$ is $6.36 	imes 10^{-3}	ext{ s}^{-1}$ and activation energy is $209	ext{ kJ mol}^{-1}$. Its rate constant (in $	ext{s}^{-1}$) at $600	ext{ K}$ is $x 	imes 10^{-6}$. The value of $x$ is .......... .
(Nearest integer)
[Given, $R = 8.31	ext{ J K}^{-1}	ext{mol}^{-1}$, $\log 6.36 	imes 10^{-3} = -2.19, 10^{-4.79} = 1.62 	imes 10^{-5}$]

Sigma Percentile · Accepted Answer

### The Arrhenius Connection

Imagine you are trying to bake a cake. If the oven is too cool, the cake takes forever to bake. If it's hot, it bakes quickly. In chemistry, the **Arrhenius equation** is the mathematical recipe that tells us exactly how the speed of a reaction (the rate constant, $k$) changes with temperature ($T$).

When we are given the rate constant at one temperature and need to find it at another, we use the two-temperature logarithmic form of the Arrhenius equation:

$$ \log \left( \frac{k_2}{k_1} ight) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} ight) $$

This elegant equation isolates the effect of temperature change, allowing us to predict the new rate constant if we know the **activation energy** ($E_a$), which is the energy barrier the molecules must overcome to react.

### Navigating the Units

Before we plug in the numbers, we must address a classic trap: **unit mismatch**.

The universal gas constant $R$ is given as $8.31 	ext{ J K}^{-1} 	ext{ mol}^{-1}$. However, our activation energy $E_a$ is given in kilo-Joules ($209 	ext{ kJ mol}^{-1}$). If we substitute these directly, our calculation will be off by a factor of a thousand!

We must convert $E_a$ to Joules:
$$ E_a = 209 	ext{ kJ mol}^{-1} = 209000 	ext{ J mol}^{-1} $$

Now, our units are perfectly aligned, and we can proceed with confidence.

### The Mathematical Crunch

Let's substitute our known values into the equation. We know $T_1 = 700 	ext{ K}$, $k_1 = 6.36 	imes 10^{-3} 	ext{ s}^{-1}$, and $T_2 = 600 	ext{ K}$.

$$ \log k_2 - \log (6.36 	imes 10^{-3}) = \frac{209000}{2.303 	imes 8.31} \left( \frac{1}{700} - \frac{1}{600} ight) $$

Notice the temperature term: $\left( \frac{1}{700} - \frac{1}{600} ight)$. Since $600 < 700$, $\frac{1}{600} > \frac{1}{700}$, making this term negative. This makes perfect physical sense! A drop in temperature means less thermal energy, fewer successful collisions, and a smaller rate constant.

Simplifying the right side:
$$ \frac{209000}{19.138} 	imes \left( \frac{600 - 700}{420000} ight) = 10920.7 	imes \left( \frac{-100}{420000} ight) = -2.60 $$

We are given that $\log (6.36 	imes 10^{-3}) = -2.19$. Substituting this into the left side:
$$ \log k_2 - (-2.19) = -2.60 $$
$$ \log k_2 + 2.19 = -2.60 $$
$$ \log k_2 = -4.79 $$

### The Final Reveal

We have $\log k_2 = -4.79$. To find $k_2$, we need to take the antilog:
$$ k_2 = 10^{-4.79} $$

The problem generously provides the value for this: $10^{-4.79} = 1.62 	imes 10^{-5}$.

So, our new rate constant is $1.62 	imes 10^{-5} 	ext{ s}^{-1}$.

The question asks for the answer in the format $x 	imes 10^{-6}$. Let's adjust our decimal point:
$$ 1.62 	imes 10^{-5} = 16.2 	imes 10^{-6} $$

Comparing this to $x 	imes 10^{-6}$, we find $x = 16.2$.

Rounding to the nearest integer, we get our final, triumphant answer: **16**.

Enter Numerical Value:

Visualized Solution

$Arrhenius Equation$

$Identifying Given Values$

$Substitution$

$Simplification$

$Calculating lo g k_{2}$

$Finding k_{2}$

$Final Answer$

The Sigma Insight: Theories of Chemical Reaction

The Arrhenius Connection

Navigating the Units

The Mathematical Crunch

The Final Reveal

Similar Questions

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 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) $[R = 8.314 J K^{- 1} mol^{- 1}]$

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; $R = 8.314 J mol^{- 1} K^{- 1}$ , $ln 3.555 = 1.268$ )

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)

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, $R = 8.31 J K^{- 1} mol^{- 1}$ ].

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

For a reaction, consider the plot of $ln k$ versus $1/ T$ given in the figure. If the rate constant of this reaction at $400 K$ is $1 0^{- 5} s^{- 1}$ , then the rate constant at $500 K$ is

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 $lo g 2 = 0.301$ )

Enter Numerical Value:

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating logk2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding k2​

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