The rate constant for the decomposition of $N_2O_5$ at various temperatures is given below: | T/°C | 0 | 20 | 40 | 60 | 80 | |---|---|---|---|---|---| | $10^5\times k$/$\text{s}^{-1}$ | 0.0787 | 1.70 | 25.7 | 178 | 2140 | Draw a graph between $\ln k$ and $1/T$ and calculate the values of $A$ and $E_a$. Predict the rate constant at 30° and 50°C.

The rate constant for the decomposition of N₂O₅ at – Class 12 Chemistry

Exercises · 3.22

The rate constant for the decomposition of $N_2O_5$ at various temperatures is given below:

T/°C	0	20	40	60	80
$10^5\times k$ / $\text{s}^{-1}$	0.0787	1.70	25.7	178	2140

Draw a graph between $\ln k$ and $1/T$ and calculate the values of $A$ and $E_a$ . Predict the rate constant at 30° and 50°C.

Telangana TsbieTextbookSubjective· 5mImportance★★★★★

35% · 41/117 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Arrhenius Equation

The Arrhenius Equation Plot: Why Temperature Changes Reaction Speed

You already know that heating things up makes reactions go faster. A cold chai takes forever to dissolve sugar; hot chai does it in seconds. But how much faster? And is there a pattern that holds for every reaction?

That pattern is the Arrhenius equation, and plotting it in a clever way reveals something fundamental about how molecules need to collide to react.

The core idea: an energy barrier

Imagine a ball sitting in a valley. To get to the next valley, it must first be pushed up over a hill. That hill is the activation energy ( $E_a$ ) — the minimum energy two molecules need to have when they collide, for the reaction to happen.

At a low temperature, most molecules move slowly. Only a tiny fraction have enough energy to climb that hill. Raise the temperature, and suddenly many more molecules have the required energy. The fraction of molecules with energy $\ge E_a$ is given by the Boltzmann distribution:

\text{fraction} = e^{-E_a / RT}

where $R$ is the gas constant and $T$ is the absolute temperature (in Kelvin). This exponential is the heart of the story.

The Arrhenius equation (precise statement)

The rate constant $k$ of a reaction depends on temperature as:

k = A \, e^{-E_a / RT}

$k$ = rate constant (how fast the reaction proceeds)
$A$ = pre-exponential factor (frequency of collisions, times a steric factor — how often molecules hit in the right orientation)
$E_a$ = activation energy (J/mol or kJ/mol)
$R$ = 8.314 J/(mol·K)
$T$ = temperature in Kelvin

Important

$k$ is not the reaction rate itself — it's the proportionality constant in the rate law. But for a fixed concentration, a larger $k$ means a faster reaction.

Why plot it? The linear trick

The equation $k = A e^{-E_a / RT}$ is exponential in $1/T$ . That's hard to eyeball. But take the natural logarithm of both sides:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is the equation of a straight line:

y = c + m x

where:

$y = \ln k$
$x = 1/T$
slope $m = -E_a / R$
intercept $c = \ln A$

So if you measure $k$ at several temperatures and plot $\ln k$ versus $1/T$ , you get a straight line — provided the reaction follows Arrhenius behaviour (most do, over moderate temperature ranges).

Tip

Always use Kelvin for $T$ . Celsius will give you a curved mess because $1/T$ is not linear in Celsius.

What the plot tells you

From the slope, you get $E_a$ :

E_a = - (\text{slope}) \times R

A steep negative slope means a large $E_a$ — the reaction is very sensitive to temperature. A shallow slope means a small $E_a$ — temperature doesn't affect it much.

From the intercept, you get $A$ :

A = e^{\text{intercept}}

This tells you about the collision frequency and orientation factor. A high $A$ means molecules are colliding often and in the right geometry.

A typical Arrhenius plot looks like this

$T$ (K)	$k$ (s⁻¹)	$1/T$ (K⁻¹)	$\ln k$
300	0.0012	0.00333	-6.72
310	0.0028	0.00323	-5.88
320	0.0061	0.00313	-5.10
330	0.0125	0.00303	-4.38

Plot $\ln k$ (y-axis) vs $1/T$ (x-axis). The points fall on a straight line. Draw the best-fit line, measure its slope, and compute $E_a$ . …

Why this formula?

Arrhenius Equation: Why It Holds

The Arrhenius equation is not a guess — it emerges from a deep physical picture of how molecules react. Let's build that picture step by step.

1. The Core Question

Why does reaction rate increase dramatically with temperature?

For many reactions, a 10 °C rise can double or triple the rate. This cannot be explained by simple kinetic energy arguments alone — the relationship is exponential.

2. The Key Insight: An Energy Barrier

Before molecules can react, they must collide — but not all collisions lead to products.

There is a minimum energy threshold called activation energy ( $E_a$ ). Only collisions with energy ≥  $E_a$ can break old bonds and form new ones.

Think of it like pushing a boulder over a hill:

The hilltop is the transition state (activated complex).
$E_a$ is the height of that hill from the reactant valley.

3. The Boltzmann Factor — The "Why" of Exponential Dependence

At temperature $T$ , the fraction of molecules with energy ≥  $E_a$ is given by the Boltzmann distribution:

\text{Fraction} = e^{-E_a / (RT)}

where:

$R$ = universal gas constant (8.314 J mol⁻¹ K⁻¹)
$T$ = absolute temperature (Kelvin)

Why this form?

The Boltzmann distribution tells us that the probability of a molecule having energy $E$ is proportional to $e^{-E/(k_B T)}$ .
For 1 mole of molecules, $k_B$ (Boltzmann constant) becomes $R$ via $R = N_A k_B$ .
So the fraction with energy ≥ $E_a$ is the integral of that distribution from $E_a$ to ∞, which yields $e^{-E_a/(RT)}$ .

Key takeaway: This exponential factor is not arbitrary — it comes directly from statistical mechanics.

4. The Pre-Exponential Factor ( $A$ )

Even if a collision has enough energy, it must also:

Have the correct orientation (steric factor)
Occur with sufficient collision frequency

These are bundled into the pre-exponential factor $A$ (also called the frequency factor):

A = \text{(collision frequency)} \times \text{(orientation factor)}

For simple gas-phase reactions, collision frequency can be calculated from kinetic molecular theory — it's on the order of $10^{10}$  L mol⁻¹ s⁻¹.

5. Putting It Together: The Arrhenius Equation

The rate constant $k$ is proportional to:

The number of effective collisions per second (given by $A$ )
The fraction of collisions with sufficient energy (given by $e^{-E_a/(RT)}$ )

Thus:

\boxed{k = A \, e^{-E_a / (RT)}}

This is the Arrhenius equation.

6. Why It Works — The Physical Logic

Component	Physical meaning	Why it's there
$A$	Maximum possible rate if every collision worked	Accounts for collision frequency & geometry

Start your 14-day free trial to unlock the full solution →

The key idea is the Arrhenius equation in its linear form:

$\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$ .

A plot of $\ln k$ vs $1/T$ gives a straight line with slope $= -E_a/R$ and intercept $= \ln A$ .

Step 1 – Convert data

Convert $T$ to Kelvin: $T/K = t/°C + 273.15$ . The given $k$ values are $10^5 \times k$ , so the actual $k$ used for $\ln k$ is the table value $\times 10^{-5}$ .

$T/°C$	$T/K$	$10^3/T$ (K $^{-1}$ )	actual $k$ (s $^{-1}$ )	$\ln k$
0	273.15	3.661	$7.87\times10^{-7}$	$-14.055$
20	293.15	3.411	$1.70\times10^{-5}$	$-10.982$
40	313.15	3.193	$2.57\times10^{-4}$	$-8.266$
60	333.15	3.002	$1.78\times10^{-3}$	$-6.331$
80	353.15	2.832	$2.14\times10^{-2}$	$-3.844$

Step 2 – Plot and find slope

Plot $\ln k$ (y-axis) vs $10^3/T$ (x-axis). Using the first and last points:

Slope $= \frac{(-3.844) - (-14.055)}{(2.832 - 3.661)\times 10^{-3}} = \frac{10.211}{-0.829 \times 10^{-3}} \approx -1.232 \times 10^4$ K.

Step 3 – Calculate $E_a$ and $A$

$E_a = -\text{slope} \times R = 1.232 \times 10^4 \times 8.314 \approx 1.024 \times 10^5$ J/mol = 102.4 kJ/mol.

Intercept $\ln A = \ln k + \frac{E_a}{R}\cdot\frac{1}{T}$ . Using the (central) point at 40°C:

$\ln A = -8.266 + (1.232\times 10^4)(3.193\times 10^{-3}) = -8.266 + 39.34 = 31.07$

$A = e^{31.07} \approx 3.1 \times 10^{13}$ s $^{-1}$ . …

Start your 14-day free trial to unlock the full solution →

Using the Arrhenius equation $\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$ , we plot $\ln k$ vs $1/T$ to get a straight line. From its slope ( $-E_a/R$ ) and intercept ( $\ln A$ ), we find $E_a \approx 102.4\ \text{kJ mol}^{-1}$ and $A \approx 3.1\times 10^{13}\ \text{s}^{-1}$ . Then we predict $k_{30^\circ\text{C}} \approx 7.0\times 10^{-5}\ \text{s}^{-1}$ and $k_{50^\circ\text{C}} \approx 8.6\times 10^{-4}\ \text{s}^{-1}$ .

The Arrhenius equation is the backbone of temperature-dependent kinetics. It tells us that the rate constant $k$ depends exponentially on temperature:

k = A e^{-E_a/RT}

Taking natural logs gives a linear form:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is of the form $y = mx + c$ , where $y = \ln k$ , $x = 1/T$ , slope $m = -E_a/R$ , and intercept $c = \ln A$ . So if we plot $\ln k$ against $1/T$ , we get a straight line — and from its slope and intercept we can extract both $E_a$ and $A$ .

Watch out

Temperature must be in kelvin when using $1/T$ in the Arrhenius plot. A common mistake is to use Celsius directly — that gives a completely wrong slope.

Let’s work through it step by step.

1. Convert temperatures to kelvin and compute $1/T$ and $\ln k$

$T$ (°C)	$T$ (K)	$1/T$ (K $^{-1}$ )	$k$ (s $^{-1}$ )	$\ln k$
0	273.15	$3.661 \times 10^{-3}$	$0.0787 \times 10^{-5}$	$-14.055$
20	293.15	$3.411 \times 10^{-3}$	$1.70 \times 10^{-5}$	$-10.982$
40	313.15	$3.193 \times 10^{-3}$	$25.7 \times 10^{-5}$	$-8.266$
60	333.15	$3.002 \times 10^{-3}$	$178 \times 10^{-5}$	$-6.331$
80	353.15	$2.832 \times 10^{-3}$	$2140 \times 10^{-5}$	$-3.844$

Notice that $k$ values are given as $10^5 \times k$ , so we divide by $10^5$ to get actual $k$ in s $^{-1}$ .

2. Plot $\ln k$ vs $1/T$

On a graph, the points fall beautifully on a straight line. The slope is negative (since $k$ increases with $T$ , $\ln k$ increases as $1/T$ decreases). We can calculate the slope using any two well-separated points, but for accuracy, use the first and last:

\text{slope} = \frac{\Delta (\ln k)}{\Delta (1/T)} = \frac{(-3.844) - (-14.055)}{(2.832 - 3.661) \times 10^{-3}} = \frac{10.211}{-0.829 \times 10^{-3}} \approx -1.232 \times 10^4\ \text{K}

Tip

Using the two extreme points gives a quick estimate. For exam problems, this is usually sufficient — but if you have time, a least-squares fit (or averaging slopes from multiple pairs) gives a more reliable result.

3. Calculate $E_a$ from the slope

Since slope $= -E_a/R$ , we have:

-\frac{E_a}{R} = -1.232 \times 10^4\ \text{K}

E_a = 1.232 \times 10^4 \times R = 1.232 \times 10^4 \times 8.314\ \text{J mol}^{-1}

E_a \approx 1.024 \times 10^5\ \text{J mol}^{-1} = 102.4\ \text{kJ mol}^{-1}

4. Calculate $A$ from the intercept

The intercept $c = \ln A$ . From the graph, the line crosses the $\ln k$ axis at $1/T = 0$ (theoretical). Using the point-slope form with any data point, say at $T = 40^\circ$ C:

\ln A = \ln k + \frac{E_a}{R} \cdot \frac{1}{T}

\ln A = -8.266 + (1.232 \times 10^4) \times (3.193 \times 10^{-3})

\ln A = -8.266 + 39.34 = 31.07

So:

$A = e^{31.07} \approx 3.1 \times 10^{13}\ \text{s}^{-1}$ …

Start your 14-day free trial to unlock the full solution →

Method: Graphical Arrhenius Analysis (Two-Point & Linear Regression)

The Arrhenius equation in logarithmic form is:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is a straight line: $y = mx + c$ , where:

$y = \ln k$
$x = 1/T$ (in Kelvin)
Slope $m = -E_a/R$
Intercept $c = \ln A$

Step 1: Convert temperatures to Kelvin and compute $1/T$ and $\ln k$

The given $k$ values in the table are $10^5 \times k$ , so the actual $k$ used to compute $\ln k$ is the table value $\times 10^{-5}$ .

$T$ (°C)	$T$ (K)	$1/T$ (K $^{-1}$ )	$k \times 10^5$ (s $^{-1}$ )	actual $k$ (s $^{-1}$ )	$\ln k$
0	273.15	$3.661 \times 10^{-3}$	0.0787	$7.87\times10^{-7}$	$-14.055$
20	293.15	$3.411 \times 10^{-3}$	1.70	$1.70\times10^{-5}$	$-10.982$
40	313.15	$3.193 \times 10^{-3}$	25.7	$2.57\times10^{-4}$	$-8.266$
60	333.15	$3.002 \times 10^{-3}$	178	$1.78\times10^{-3}$	$-6.331$
80	353.15	$2.832 \times 10^{-3}$	2140	$2.14\times10^{-2}$	$-3.844$

Step 2: Plot $\ln k$ (y-axis) vs $1/T$ (x-axis)

You will get a straight line with a negative slope.

Step 3: Calculate slope from the graph

Using the first and last points (for a quick estimate):

\text{slope} = \frac{\Delta \ln k}{\Delta (1/T)} = \frac{-3.844 - (-14.055)}{(2.832 - 3.661) \times 10^{-3}}

= \frac{10.211}{-0.829 \times 10^{-3}} \approx -1.232 \times 10^4 \text{ K}

Step 4: Calculate activation energy $E_a$

From slope $m = -E_a/R$ :

E_a = -m \times R = 1.232 \times 10^4 \times 8.314

\boxed{E_a \approx 1.024 \times 10^5 \text{ J/mol} = 102.4 \text{ kJ/mol}}

Step 5: Calculate pre-exponential factor $A$

From intercept $c = \ln A$ :

Using the central point at $T = 313.15$ K ( $1/T = 3.193 \times 10^{-3}$ , $\ln k = -8.266$ ):

\ln A = \ln k + \frac{E_a}{R} \cdot \frac{1}{T}

\ln A = -8.266 + (1.232 \times 10^4)(3.193 \times 10^{-3})

\ln A = -8.266 + 39.34 = 31.07

\boxed{A = e^{31.07} \approx 3.1 \times 10^{13} \text{ s}^{-1}}

Step 6: Predict rate constants at 30°C and 50°C

Use the fitted line $\ln k = 31.07 - \dfrac{1.232 \times 10^4}{T}$ .

At 30°C (303.15 K):

$\ln k = 31.07 - (1.232 \times 10^4)(3.299 \times 10^{-3})$ …

Start your 14-day free trial to unlock the full solution →

Common Mistakes & How to Avoid Them (Arrhenius Equation)

1. ✗ Forgetting to convert temperature to Kelvin

The Mistake: Students plot $1/T$ using °C values directly (e.g., $1/20$ instead of $1/293$ ).

Why it's wrong: The Arrhenius equation uses absolute temperature:

k = A e^{-E_a/RT}

$T$ must be in Kelvin ( $K = °C + 273$ ).

✓ How to avoid: Always write the conversion step explicitly:

$0°C = 273\,K$
$20°C = 293\,K$
$40°C = 313\,K$ , etc.

2. ✗ Using $k$ directly instead of $\ln k$

The Mistake: Plotting $k$ vs $1/T$ (a curve) instead of $\ln k$ vs $1/T$ (a straight line).

Why it's wrong: The Arrhenius equation in linear form is:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

Only $\ln k$ vs $1/T$ gives a straight line with slope $= -E_a/R$ .

✓ How to avoid: Before plotting, compute $\ln k$ for each $k$ value. Use a table:

$T/K$	$10^5 k$	$\ln k$	$1/T$
273	0.0787	$\ln(0.0787 \times 10^{-5})$	0.00366
...	...	...	...

3. ✗ Mishandling the $10^5$ factor in $k$

The Mistake: Taking $\ln(0.0787)$ instead of $\ln(0.0787 \times 10^{-5})$ .

Why it's wrong: The given $k$ values are $10^5 \times k$ . So actual $k = (\text{table value}) \times 10^{-5}$ .

✓ How to avoid: Write clearly:

k_{\text{actual}} = (\text{table value}) \times 10^{-5}

Then take $\ln$ of this actual value.

4. ✗ Using $R$ in wrong units

The Mistake: Using $R = 0.0821$ (L·atm/mol·K) instead of $R = 8.314$ (J/mol·K).

Why it's wrong: $E_a$ is typically in J/mol or kJ/mol. The correct $R$ for energy calculations is:

R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}

✓ How to avoid: Remember:

For $E_a$ in J/mol: use $R = 8.314$
For $E_a$ in kJ/mol: use $R = 0.008314$

5. ✗ Confusing slope sign when finding $E_a$

The Mistake: Taking $E_a = \text{slope} \times R$ instead of $E_a = -\text{slope} \times R$ .

Why it's wrong: From $\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$ , the slope is negative:

\text{slope} = -\frac{E_a}{R}

So $E_a = -\text{slope} \times R$ (which gives a positive value).

✓ How to avoid:

Plot the graph
Calculate slope $= \frac{\Delta \ln k}{\Delta (1/T)}$ (will be negative)
Then $E_a = -\text{slope} \times R$

6. ✗ Using wrong points for interpolation at 30°C and 50°C

The Mistake: Reading $k$ directly from the curved $k$ vs $T$ plot.

Why it's wrong: The relationship is linear only for $\ln k$ vs $1/T$ . …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$Q. At what temperature will the RMS velocity of sulphur dioxide molecules at 400 K be the same as the most probable velocity of oxygen molecules? (A) 600 K (B) 200 K (C) 400 K (D) 300 K$
›Reveal solutionSolution

Equating $v_{rms}(\text{SO}_2, 400\,\text{K})$ with $v_{mp}(\text{O}_2, T)$ gives $T = 300$ K.

$v_{rms} = \sqrt{\frac{3RT_1}{M_1}}, \qquad v_{mp} = \sqrt{\frac{2RT_2}{M_2}}.$
With $M(\text{SO}_2) = 64$ , $T_1 = 400$ K, and $M(\text{O}_2) = 32$ , set the two speeds equal:

$\frac{3RT_1}{M_1} = \frac{2RT_2}{M_2}$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$^\circ$
›Reveal solutionSolution
The temperature coefficient (Q₁₀ = 2) tells us the rate constant halves for every 10 °C drop. From 25 °C to 15 °C is a 10 °C decrease, so the rate constant at 15 °C is half of $1 \times 10^{-3}$ min⁻¹, i.e., $5 \times 10^{-4}$ min⁻¹.

Concept & Intuition

The temperature coefficient (often denoted $Q_{10}$ ) is defined as the factor by which the rate constant increases when the temperature is raised by 10 °C.

Here $Q_{10} = 2$ means:

Increase temperature by 10 °C → rate constant doubles.

Decrease temperature by 10 °C → rate constant halves.

We are moving down from 25 °C to 15 °C, a drop of exactly 10 °C. So the rate constant at the lower temperature is simply $k_{15} = k_{25} / 2$ .

Step-by-step reasoning

Identify the given data

$k_{25} = 1 \times 10^{-3}$ min⁻¹

Temperature coefficient $Q_{10} = 2$

Temperature change: from 25 °C to 15 °C = −10 °C

Apply the definition of temperature coefficient

The relationship is:

$\frac{k_{T+10}}{k_T} = Q_{10}$
So for a 10 °C decrease:
$\frac{k_{15}}{k_{25}} = \frac{1}{Q_{10}} = \frac{1}{2}$

Calculate the unknown rate constant $k_{15} = k_{25} \times \frac{1}{2} = (1 \times 10^{-3}) \times \frac{1}{2} = 0.5 \times 10^{-3} = 5 \times 10^{-4} \text{ min}^{-1}$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$25^\circ \mathrm{C}$
›Reveal solutionSolution
The temperature coefficient (Q₁₀ = 2) tells us the rate constant halves for every 10°C drop. Going from 25°C to 15°C is a 10°C decrease, so the rate constant at 15°C is half of the given value: $5 \times 10^{-4}~\mathrm{min}^{-1}$ .

Concept & Intuition

The temperature coefficient (often denoted $Q_{10}$ ) is the factor by which the rate constant changes when the temperature is raised by 10°C. Here $Q_{10} = 2$ means that for every 10°C increase, the rate constant doubles; conversely, for every 10°C decrease, it halves. Since we are moving from 25°C down to 15°C (a drop of exactly 10°C), we simply divide the given rate constant by 2.

Step-by-step reasoning

Identify the temperature change

The given rate constant $k_{25} = 1 \times 10^{-3}~\mathrm{min}^{-1}$ is at $25^\circ\mathrm{C}$ . We need $k_{15}$ at $15^\circ\mathrm{C}$ .

The difference: $25^\circ\mathrm{C} - 15^\circ\mathrm{C} = 10^\circ\mathrm{C}$ .

Apply the definition of the temperature coefficient

The temperature coefficient $Q_{10}$ is defined as

$Q_{10} = \frac{k_{T+10}}{k_T}$
Here $Q_{10} = 2$ , so for a 10°C increase the rate constant multiplies by 2. For a 10°C decrease, we use the reciprocal:
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
$Q. When salt is added to water, which of the following statement is true? (A) Boiling point decreases (B) Boiling point increases (C) Boiling point remain constant (D) Freezing point increases$
›Reveal solutionSolution
Adding salt to water raises its boiling point and lowers its freezing point — a colligative effect. The correct statement is that the boiling point increases.

The question tests your understanding of colligative properties — properties that depend only on the number of solute particles, not their chemical identity. When salt (sodium chloride, NaCl) dissolves in water, it dissociates into Na⁺ and Cl⁻ ions, increasing the total number of particles in the solution. This changes two key physical properties of water: its boiling point and its freezing point.

Why does this happen? At the boiling point, the vapour pressure of the liquid equals the atmospheric pressure. Adding a non-volatile solute like salt lowers the vapour pressure of the solvent (water). To make the vapour pressure reach atmospheric pressure again, you need to supply more heat — hence the boiling point rises. For freezing, the solute particles disrupt the orderly arrangement of water molecules into ice, so a lower temperature is needed to freeze the solution — hence the freezing point drops.

Let’s examine each option step by step.

Boiling point behaviour

The boiling point elevation is given by $\Delta T_b = i \cdot K_b \cdot m$ , where $i$ is the van’t Hoff factor (for NaCl, $i \approx 2$ ), $K_b$ is the ebullioscopic constant of water, and $m$ is the molality. Since $\Delta T_b > 0$ , the boiling point increases. This eliminates option (A) and (C).

Freezing point behaviour …
Start your 14-day free trial to see the full solution →

←Previous question3.21 — The following data were obtained during the first order thermal decomp…→Next question3.23 — The rate constant for the decomposition of hydrocarbons is 2.418×10⁻⁵s…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days