Why in the redox titration of $KMnO_4$ vs oxalic acid, we heat oxalic acid solution before starting the titration?

Why in the redox titration of KMnO₄ vs oxalic acid, we heat oxalic acid solution before starting the titration

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KMnO_4

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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

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The reaction between $KMnO_4$ and oxalic acid is slow at room temperature because the activation energy barrier is high. Heating the oxalic acid solution provides the necessary kinetic energy to overcome this barrier, ensuring the titration proceeds at a measurable and accurate rate.

The Arrhenius equation is the key to understanding this. It tells us that the rate constant $k$ of a reaction depends exponentially on temperature:

k = A e^{-E_a / RT}

Here, $E_a$ is the activation energy — the minimum energy colliding molecules must have for a reaction to occur. For the reaction between permanganate ions ( $MnO_4^-$ ) and oxalic acid ( $H_2C_2O_4$ ), $E_a$ is quite large. At room temperature, very few collisions have enough energy, so the reaction is painfully slow — you'd be waiting minutes for a single drop of $KMnO_4$ to decolourise. Heating dramatically increases the fraction of molecules with energy $\ge E_a$ , speeding up the reaction to a practical rate.

Let's walk through the titration logic step by step.

The reaction itself is autocatalytic. The balanced equation in acidic medium is:

2MnO_4^- + 5H_2C_2O_4 + 6H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O

The $Mn^{2+}$ ions produced act as a catalyst for the further reaction. So initially, even at room temperature, the reaction is extremely slow. Once a little $Mn^{2+}$ forms, it speeds up — but that initial lag makes the endpoint unclear and the titration inaccurate.

Heating solves the initial lag. By warming the oxalic acid solution to about 60–70 °C (never boiling, as oxalic acid can decompose), we supply enough thermal energy to overcome the activation barrier right from the start. The first few drops of $KMnO_4$ react quickly, producing $Mn^{2+}$ immediately. The autocatalytic effect then kicks in, and the reaction proceeds briskly throughout the titration.
Why not heat the $KMnO_4$ solution? $KMnO_4$ is thermally unstable. Heating it would cause it to decompose:

2KMnO_4 \xrightarrow{\Delta} K_2MnO_4 + MnO_2 + O_2

This changes the concentration of the titrant and introduces $MnO_2$ (a brown precipitate), ruining the titration. So we only heat the oxalic acid solution. …

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Here are the common mistakes students make when answering why oxalic acid is heated before titration with $KMnO_4$ , along with how to avoid each.

Mistake 1: Saying "to increase the rate of reaction" without specifying the slow step

Why it’s wrong:

This is too vague. Many reactions speed up on heating — the examiner wants the specific chemical reason.

How to avoid:

Always mention that the reaction between $KMnO_4$ and oxalic acid is autocatalytic (catalysed by $Mn^{2+}$ ions produced). Initially, the reaction is very slow because $Mn^{2+}$ is absent. Heating provides the activation energy to start the reaction, generating the first $Mn^{2+}$ ions, which then catalyse the rest.

Key point: Heat is needed to initiate the autocatalytic cycle, not just to speed up a fast reaction.

Mistake 2: Confusing the role of heat with decomposition of oxalic acid

Why it’s wrong:

Oxalic acid decomposes on strong heating (above ~150°C) to give $CO_2$ , $CO$ , and water. But in titration, we only warm (50–60°C), not boil. Students often write that we heat to "decompose oxalic acid" — which would ruin the titration.

How to avoid:

Clearly state: We warm the solution to around 50–60°C (not boiling). This is enough to overcome the activation energy barrier without decomposing the oxalic acid.

Mistake 3: Forgetting the autocatalytic role of $Mn^{2+}$

Why it’s wrong:

Some students say "heat increases kinetic energy of molecules" — which is true but incomplete. The real mechanism involves $Mn^{2+}$ as a catalyst.

How to avoid:

Write the two-step mechanism in your answer:

Slow step (without catalyst):

$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$

(very slow initially)
Fast step (catalysed by $Mn^{2+}$ ):

$Mn^{2+}$ formed in step 1 catalyses further reaction, making it rapid.

Heat is needed to initiate step 1 so that step 2 can take over.

Mistake 4: Thinking heat is needed throughout the titration

Why it’s wrong:

Once the reaction starts and $Mn^{2+}$ builds up, the reaction becomes fast even without continued heating. Students sometimes keep the flask on a hot plate the whole time.

How to avoid:

Explain that heating is only required at the start. After the first few drops of $KMnO_4$ decolourise, the autocatalysis sustains the rate. In fact, continued heating might cause oxalic acid decomposition or loss of volume by evaporation.

Mistake 5: Not mentioning the colour change as evidence

Why it’s wrong: …

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

Showing the 12 most recent of 13 on this concept.

AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$\exp\left(-\dfrac{E_a}{RT}\right)$
›Reveal solutionSolution
This tests the meaning of each term in the Arrhenius equation; the exponential term represents the fraction of molecules with energy exceeding the activation energy.

Concept and Intuition

The Arrhenius equation, $k = A e^{-E_a/RT}$ , splits the rate constant into two physically distinct factors: $A$ (the frequency/pre-exponential factor, related to collision frequency and orientation) and $e^{-E_a/RT}$ (the fraction of molecular collisions that have enough energy to cross the activation energy barrier, derived from the Maxwell-Boltzmann energy distribution). Only molecules with energy $\geq E_a$ can react upon collision; the exponential term quantifies what fraction of the population meets this threshold at a given temperature.

Step-by-Step Solution

Write the Arrhenius equation: $k = A\,e^{-E_a/RT}$ .

Recognize $A$ is the frequency factor — related to the rate of collisions and their proper orientation, independent of energy considerations.

Recognize $e^{-E_a/RT}$ arises from integrating the Maxwell-Boltzmann distribution above the energy threshold $E_a$ — it represents the fraction of molecules whose energy exceeds $E_a$ at temperature $T$ . …
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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$2.5 \times 10^{-4} s^{-1}$
›Reveal solutionSolution
This tests the Arrhenius equation relating rate constants at two temperatures; solving gives $T_2 \approx 625\,K$ .

Concept and Intuition

A faster half-life means a larger rate constant. The Arrhenius equation connects the ratio of two rate constants at two temperatures to the activation energy — a bigger $E_a$ needs a bigger temperature jump to boost $k$ by the same factor.

Step-by-Step Solution

$T_1 = 300°C = 573\,K$ , $k_1 = 2.5\times10^{-4}\,s^{-1}$ .

Desired half-life $t_{1/2}=138.6\,s$ for a first-order reaction: $k_2 = \dfrac{0.693}{138.6} = 5\times10^{-3}\,s^{-1}$ .

Ratio: $k_2/k_1 = \dfrac{5\times10^{-3}}{2.5\times10^{-4}} = 20$ .

Arrhenius (two-temperature form): $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ .

$\dfrac{E_a}{2.303R} = \dfrac{42000}{2.303\times2} = 9118.5$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$1.6\times10^{-5}s^{-1}$
›Reveal solutionSolution
The Arrhenius equation converts rate constants at two temperatures via activation energy; the given antilog hint (antilog(0.4771)=3.0) is a shortcut for the final power-of-ten step, giving $k_2=4.8\times10^{-3}\ s^{-1}$ .

Concept and Intuition

The two-point form of the Arrhenius equation lets us find how much a rate constant changes when temperature changes, purely from the activation energy — a higher $E_a$ makes the rate far more sensitive to temperature. The problem hands us the exact antilog we'll need at the end, which is a strong signal the intended computation lands exactly on $\log(k_2/k_1)=2.4771$ .

Step-by-Step Solution

Write the two-temperature Arrhenius relation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

Compute $\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{600}-\dfrac{1}{700} = \dfrac{700-600}{600\times700} = \dfrac{100}{420000} = \dfrac{1}{4200}$ .

Compute $\dfrac{E_a}{2.303R} = \dfrac{198870}{2.303\times8.3} = \dfrac{198870}{19.1149} \approx 10403.9$ .

Multiply: $10403.9 \times \dfrac{1}{4200} \approx 2.4771$ . …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$\to$
›Reveal solutionSolution
This tests the two-point form of the Arrhenius equation to find an unknown temperature given a rate constant ratio. The answer works out to T ≈ 311 K.

Concept and Intuition

The Arrhenius equation $k = A e^{-E_a/RT}$ can be converted, for two temperatures, into a linear relation between $\log(k_2/k_1)$ and $(1/T_1 - 1/T_2)$ , letting us solve for an unknown temperature when the rate constants and activation energy are known.

Step-by-Step Solution

Two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303 R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

Given $k_1 = 5\times10^3\ s^{-1}$ at $T_1 = 300\ K$ , and $k_2 = 1.0\times10^4\ s^{-1}$ at $T_2 = T$ (to find). Ratio $k_2/k_1 = \dfrac{1.0\times10^4}{5\times10^3} = 2$ .

Given $\log 2 = 0.3$ , so LHS $= 0.3$ .

Compute $\dfrac{E_a}{2.303R} = \dfrac{50000}{2.303 \times 8.3} = \dfrac{50000}{19.115} \approx 2615.7$ .

So $0.3 = 2615.7\left(\dfrac{1}{300} - \dfrac{1}{T}\right)$ , giving $\dfrac{1}{300} - \dfrac{1}{T} = \dfrac{0.3}{2615.7} \approx 1.1468\times10^{-4}$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$\left(\dfrac{k_{610}}{k_{600}}\right)$
›Reveal solutionSolution
Same reaction time for 10% completion at 600K and 20% completion at 610K lets the time cancel, leaving $k_{610}/k_{600}=\log(1.25)/\log(1.111)\approx2.12$ .

Concept and Intuition

For a first-order reaction, $k=\dfrac{1}{t}\ln\dfrac{1}{1-x}$ where $x$ is the fraction reacted. If the same time $t$ produces different completions at two temperatures, that common $t$ cancels out when taking the ratio of the two rate constants, leaving a ratio of two logarithmic (completion) factors only.

Step-by-Step Solution

At 600 K, 10% complete in time $t$ : $k_{600} = \dfrac{1}{t}\ln\dfrac{1}{0.90} = \dfrac{2.303}{t}\log(1.111)$ .

At 610 K, 20% complete in the same time $t$ : $k_{610} = \dfrac{1}{t}\ln\dfrac{1}{0.80} = \dfrac{2.303}{t}\log(1.25)$ .

Ratio: $\dfrac{k_{610}}{k_{600}} = \dfrac{\log(1.25)}{\log(1.111)} = \dfrac{0.0969}{0.0457} \approx 2.12$ . …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$E_a$
›Reveal solutionSolution
Statement I is correct because reaction orders are experimentally determined and can be zero, positive integers, or fractions. Statement II is incorrect because the frequency factor (pre-exponential factor) is not a fraction of molecules with energy above $E_a$ ; that fraction is given by the exponential term $e^{-E_a/RT}$ . Therefore, the correct option is (C).

The Arrhenius equation is the backbone of chemical kinetics when it comes to temperature dependence. It is written as:
$k = A e^{-E_a/RT}$
Here, $k$ is the rate constant, $A$ is the frequency factor (also called the pre-exponential factor), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature. The key intuition:

$A$ represents how often molecules collide in the correct orientation (a frequency, not a fraction).

The exponential term $e^{-E_a/RT}$ is the fraction of molecules that have energy equal to or greater than $E_a$ . Mixing these two up is a classic mistake.

Now, let’s evaluate each statement carefully.

Statement I: "Order of a reaction can be obtained from experiment and can have zero or positive integer or positive fraction values"

The order of a reaction is defined as the sum of the exponents of concentration terms in the experimentally determined rate law.

It is not derived from the stoichiometric coefficients (that would be molecularity, which applies only to elementary steps).

Orders can indeed be zero (e.g., decomposition of ammonia on a platinum surface), positive integers (most common), or positive fractions (e.g., the reaction $H_2 + Br_2 \rightarrow 2HBr$ has order 1.5).

Negative orders are also possible, but the statement only claims zero, positive integers, or positive fractions — all of which are valid.

Conclusion: Statement I is correct.

Statement II: "In the Arrhenius equation, the frequency factor is the fraction of molecules that can have energy higher than $E_a$ "

Let’s revisit the Arrhenius equation: $k = A e^{-E_a/RT}$ . …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$2\times10^{-5}\ \mathrm{s^{-1}}$
›Reveal solutionSolution
Plugging the two rate constants and temperatures into the two-point Arrhenius equation gives an activation energy of about 88 kJ/mol.

Concept and Intuition

The Arrhenius equation's two-temperature form lets you find activation energy directly from two rate constants at two known temperatures, without knowing the pre-exponential factor.

Step-by-Step Solution

Two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

$\dfrac{k_2}{k_1} = \dfrac{4\times10^{-3}}{2\times10^{-5}} = 200$ ; $\log 200 = \log 2 + \log 100 = 0.3 + 2 = 2.3$ .

$\dfrac{1}{T_1} - \dfrac{1}{T_2} = \dfrac{1}{400} - \dfrac{1}{500} = \dfrac{500-400}{400\times500} = \dfrac{100}{200000} = 5\times10^{-4}\ \mathrm{K^{-1}}$ .

$2.3 = \dfrac{E_a}{2.303 \times 8.3} \times 5\times10^{-4}$ .

$2.303 \times 8.3 = 19.115$ . …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$3.46 \times 10^{-2}\ s^{-1}$
›Reveal solutionSolution
This applies the two-temperature Arrhenius equation to find the rate constant at a higher temperature given the activation energy; the result is closest to $0.692\ s^{-1}$ .

Concept and Intuition

The Arrhenius equation, $k = Ae^{-E_a/RT}$ , shows that rate constants increase sharply with temperature for reactions with a sizeable activation energy. Taking the equation at two temperatures and dividing eliminates the pre-exponential factor $A$ , giving a convenient logarithmic two-point formula for finding a rate constant at a new temperature.

Step-by-Step Solution

Write the two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

Compute $\dfrac{1}{T_1} - \dfrac{1}{T_2} = \dfrac{1}{298} - \dfrac{1}{350} = \dfrac{350-298}{298\times350} = \dfrac{52}{104300} \approx 4.986\times10^{-4}\ \text{K}^{-1}$ .

Compute $\dfrac{E_a}{2.303R} = \dfrac{50100}{2.303\times8.314} = \dfrac{50100}{19.147} \approx 2616.6$ .

Multiply: $\log\dfrac{k_2}{k_1} \approx 2616.6 \times 4.986\times10^{-4} \approx 1.305$ .

So $\dfrac{k_2}{k_1} = 10^{1.305} = 10 \times 10^{0.305}$ . Using $\log 2 = 0.3010$ , $10^{0.305}$ is just slightly above 2, i.e. $\approx 2.02$ , giving $k_2/k_1 \approx 20.2$ . …
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AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
$\mathrm{kJ\,mol^{-1}}$
›Reveal solutionSolution
Plugging the rate-doubling data into the two-point Arrhenius equation gives $E_a \approx 53.33\ \mathrm{kJ\,mol^{-1}}$ — option (D).

Concept and Intuition

The Arrhenius equation, $k = Ae^{-E_a/RT}$ , links a reaction's rate constant to temperature through its activation energy. When we know the rate constant at two temperatures, we can eliminate the pre-exponential factor $A$ and solve directly for $E_a$ using the two-temperature form of the equation — this is the standard way "activation energy from a rate-doubling" problems are solved.

Step-by-Step Solution

Two-point Arrhenius form: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ .

Here $k_2/k_1 = 2$ (rate doubled), so $\log 2 = 0.3$ (given).

$\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{300}-\dfrac{1}{310} = \dfrac{310-300}{300\times310} = \dfrac{10}{93000} = 1.0753\times10^{-4}\ \mathrm{K^{-1}}$ .

$0.3 = \dfrac{E_a}{2.303\times 8.3}\times 1.0753\times10^{-4}$ .

$2.303\times 8.3 = 19.115$ . …
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AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$k_1$
›Reveal solutionSolution
Plugging $E_a$ , $R$ , $T_1=300$ K and $T_2=400$ K into the two-point Arrhenius equation gives $\ln(k_2/k_1)\approx 4.166$ .

Concept and Intuition

The Arrhenius equation $k = A\,e^{-E_a/RT}$ shows that rate constants at two temperatures are related by $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac1{T_1}-\dfrac1{T_2}\right)$ , since the pre-exponential factor $A$ cancels out. A higher activation energy or a larger temperature gap gives a bigger ratio $k_2/k_1$ .

Step-by-Step Solution

Convert $E_a$ to joules: $41.5\ \mathrm{kJ/mol} = 41500\ \mathrm{J/mol}$ .

Compute $\dfrac{E_a}{R} = \dfrac{41500}{8.3} = 5000$ .

Compute $\dfrac1{T_1}-\dfrac1{T_2} = \dfrac1{300}-\dfrac1{400} = \dfrac{4-3}{1200} = \dfrac1{1200}$ . …
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AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$\mathrm{C_2H_5I(g) \rightarrow C_2H_4(g) + HI(g)}$
›Reveal solutionSolution
Plugging the given rate-constant ratio and temperatures into the two-point Arrhenius equation gives an activation energy of 48.16 kJ/mol.

Concept and Intuition

The Arrhenius equation relates the rate constant at two different temperatures to the activation energy: $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ . Given rate constants at two temperatures, we can solve directly for $E_a$ .

Step-by-Step Solution

$k_1 = x$ at $T_1=600$ K, $k_2=4x$ at $T_2=700$ K, so $k_2/k_1 = 4$ .

$\ln4 = 2.303\log4 = 2.303\times0.6 = 1.3818$ .

$\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac1{600}-\dfrac1{700} = \dfrac{700-600}{600\times700}=\dfrac{100}{420000}=2.381\times10^{-4}\ \text{K}^{-1}$ . …
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AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$^{-1}$
›Reveal solutionSolution
Applying the two-point Arrhenius equation gives activation energy $\approx 33.45$ kJ/mol.

Concept and Intuition

The Arrhenius equation $k = A e^{-E_a/RT}$ links rate constant to temperature through the activation energy. Taking the ratio of rate constants at two temperatures eliminates the pre-exponential factor $A$ , letting us solve directly for $E_a$ .

Step-by-Step Solution

Write $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ .

$\dfrac{k_2}{k_1} = \dfrac{0.2}{0.02}=10$ , so $\ln 10 = 2.303$ .

$\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{500}-\dfrac{1}{700} = \dfrac{700-500}{500\times700} = \dfrac{200}{350000} = 5.714\times10^{-4}\ \text{K}^{-1}$ . …
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Q.Why in the redox titration of KMnO4KMnO_4KMnO4​ vs oxalic acid, we heat oxalic acid solution before starting the titration?

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Q.Why in the redox titration of $KMnO_4$ vs oxalic acid, we heat oxalic acid solution before starting the titration?

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