The rate constant for the decomposition of N2O5 at various temperatures is given below:
| T/°C | 0 | 20 | 40 | 60 | 80 |
|---|---|---|---|---|---|
| 105×k/s−1 | 0.0787 | 1.70 | 25.7 | 178 | 2140 |
Draw a graph between lnk and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.
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🔒 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 (Ea) — 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 ≥Ea is given by the Boltzmann distribution:
fraction=e−Ea/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=Ae−Ea/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)
- Ea = activation energy (J/mol or kJ/mol)
- R = 8.314 J/(mol·K)
- T = temperature in Kelvin
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=Ae−Ea/RT is exponential in 1/T. That's hard to eyeball. But take the natural logarithm of both sides:
lnk=lnA−REa⋅T1
This is the equation of a straight line:
y=c+mx
where:
- y=lnk
- x=1/T
- slope m=−Ea/R
- intercept c=lnA
So if you measure k at several temperatures and plot lnk versus 1/T, you get a straight line — provided the reaction follows Arrhenius behaviour (most do, over moderate temperature ranges).
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 Ea:
Ea=−(slope)×R
A steep negative slope means a large Ea — the reaction is very sensitive to temperature. A shallow slope means a small Ea — temperature doesn't affect it much.
From the intercept, you get A:
A=eintercept
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) | lnk |
|---|---|---|---|
| 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 lnk (y-axis) vs 1/T (x-axis). The points fall on a straight line. Draw the best-fit line, measure its slope, and compute Ea. …
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 (Ea). Only collisions with energy ≥ Ea 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).
- Ea 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 ≥ Ea is given by the Boltzmann distribution:
Fraction=e−Ea/(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/(kBT).
- For 1 mole of molecules, kB (Boltzmann constant) becomes R via R=NAkB.
- So the fraction with energy ≥ Ea is the integral of that distribution from Ea to ∞, which yields e−Ea/(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=(collision frequency)×(orientation factor)
For simple gas-phase reactions, collision frequency can be calculated from kinetic molecular theory — it's on the order of 1010 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−Ea/(RT))
Thus:
k=Ae−Ea/(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 |
The key idea is the Arrhenius equation in its linear form:
lnk=lnA−REa⋅T1.
A plot of lnk vs 1/T gives a straight line with slope =−Ea/R and intercept =lnA.
Step 1 – Convert data
Convert T to Kelvin: T/K=t/°C+273.15. The given k values are 105×k, so the actual k used for lnk is the table value ×10−5.
| T/°C | T/K | 103/T (K−1) | actual k (s−1) | lnk |
|---|---|---|---|---|
| 0 | 273.15 | 3.661 | 7.87×10−7 | −14.055 |
| 20 | 293.15 | 3.411 | 1.70×10−5 | −10.982 |
| 40 | 313.15 | 3.193 | 2.57×10−4 | −8.266 |
| 60 | 333.15 | 3.002 | 1.78×10−3 | −6.331 |
| 80 | 353.15 | 2.832 | 2.14×10−2 | −3.844 |
Step 2 – Plot and find slope
Plot lnk (y-axis) vs 103/T (x-axis). Using the first and last points:
Slope =(2.832−3.661)×10−3(−3.844)−(−14.055)=−0.829×10−310.211≈−1.232×104 K.
Step 3 – Calculate Ea and A
Ea=−slope×R=1.232×104×8.314≈1.024×105 J/mol = 102.4 kJ/mol.
Intercept lnA=lnk+REa⋅T1. Using the (central) point at 40°C:
lnA=−8.266+(1.232×104)(3.193×10−3)=−8.266+39.34=31.07
A=e31.07≈3.1×1013 s−1. …
Using the Arrhenius equation lnk=lnA−REa⋅T1, we plot lnk vs 1/T to get a straight line. From its slope (−Ea/R) and intercept (lnA), we find Ea≈102.4 kJ mol−1 and A≈3.1×1013 s−1. Then we predict k30∘C≈7.0×10−5 s−1 and k50∘C≈8.6×10−4 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=Ae−Ea/RT
Taking natural logs gives a linear form:
lnk=lnA−REa⋅T1
This is of the form y=mx+c, where y=lnk, x=1/T, slope m=−Ea/R, and intercept c=lnA. So if we plot lnk against 1/T, we get a straight line — and from its slope and intercept we can extract both Ea and A.
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 lnk
| T (°C) | T (K) | 1/T (K−1) | k (s−1) | lnk |
|---|---|---|---|---|
| 0 | 273.15 | 3.661×10−3 | 0.0787×10−5 | −14.055 |
| 20 | 293.15 | 3.411×10−3 | 1.70×10−5 | −10.982 |
| 40 | 313.15 | 3.193×10−3 | 25.7×10−5 | −8.266 |
| 60 | 333.15 | 3.002×10−3 | 178×10−5 | −6.331 |
| 80 | 353.15 | 2.832×10−3 | 2140×10−5 | −3.844 |
Notice that k values are given as 105×k, so we divide by 105 to get actual k in s−1.
2. Plot lnk vs 1/T
On a graph, the points fall beautifully on a straight line. The slope is negative (since k increases with T, lnk increases as 1/T decreases). We can calculate the slope using any two well-separated points, but for accuracy, use the first and last:
slope=Δ(1/T)Δ(lnk)=(2.832−3.661)×10−3(−3.844)−(−14.055)=−0.829×10−310.211≈−1.232×104 K
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 Ea from the slope
Since slope =−Ea/R, we have:
−REa=−1.232×104 K
Ea=1.232×104×R=1.232×104×8.314 J mol−1
Ea≈1.024×105 J mol−1=102.4 kJ mol−1
4. Calculate A from the intercept
The intercept c=lnA. From the graph, the line crosses the lnk axis at 1/T=0 (theoretical). Using the point-slope form with any data point, say at T=40∘C:
lnA=lnk+REa⋅T1
lnA=−8.266+(1.232×104)×(3.193×10−3)
lnA=−8.266+39.34=31.07
So:
A=e31.07≈3.1×1013 s−1 …
Method: Graphical Arrhenius Analysis (Two-Point & Linear Regression)
The Arrhenius equation in logarithmic form is:
lnk=lnA−REa⋅T1
This is a straight line: y=mx+c, where:
- y=lnk
- x=1/T (in Kelvin)
- Slope m=−Ea/R
- Intercept c=lnA
Step 1: Convert temperatures to Kelvin and compute 1/T and lnk
The given k values in the table are 105×k, so the actual k used to compute lnk is the table value ×10−5.
| T (°C) | T (K) | 1/T (K−1) | k×105 (s−1) | actual k (s−1) | lnk |
|---|---|---|---|---|---|
| 0 | 273.15 | 3.661×10−3 | 0.0787 | 7.87×10−7 | −14.055 |
| 20 | 293.15 | 3.411×10−3 | 1.70 | 1.70×10−5 | −10.982 |
| 40 | 313.15 | 3.193×10−3 | 25.7 | 2.57×10−4 | −8.266 |
| 60 | 333.15 | 3.002×10−3 | 178 | 1.78×10−3 | −6.331 |
| 80 | 353.15 | 2.832×10−3 | 2140 | 2.14×10−2 | −3.844 |
Step 2: Plot lnk (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):
slope=Δ(1/T)Δlnk=(2.832−3.661)×10−3−3.844−(−14.055)
=−0.829×10−310.211≈−1.232×104 K
Step 4: Calculate activation energy Ea
From slope m=−Ea/R:
Ea=−m×R=1.232×104×8.314
Ea≈1.024×105 J/mol=102.4 kJ/mol
Step 5: Calculate pre-exponential factor A
From intercept c=lnA:
Using the central point at T=313.15 K (1/T=3.193×10−3, lnk=−8.266):
lnA=lnk+REa⋅T1
lnA=−8.266+(1.232×104)(3.193×10−3)
lnA=−8.266+39.34=31.07
A=e31.07≈3.1×1013 s−1
Step 6: Predict rate constants at 30°C and 50°C
Use the fitted line lnk=31.07−T1.232×104.
At 30°C (303.15 K):
lnk=31.07−(1.232×104)(3.299×10−3) …
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=Ae−Ea/RT
T must be in Kelvin (K=°C+273).
✓ How to avoid: Always write the conversion step explicitly:
- 0°C=273K
- 20°C=293K
- 40°C=313K, etc.
2. ✗ Using k directly instead of lnk
The Mistake: Plotting k vs 1/T (a curve) instead of lnk vs 1/T (a straight line).
Why it's wrong: The Arrhenius equation in linear form is:
lnk=lnA−REa⋅T1
Only lnk vs 1/T gives a straight line with slope =−Ea/R.
✓ How to avoid: Before plotting, compute lnk for each k value. Use a table:
| T/K | 105k | lnk | 1/T |
|---|---|---|---|
| 273 | 0.0787 | ln(0.0787×10−5) | 0.00366 |
| ... | ... | ... | ... |
3. ✗ Mishandling the 105 factor in k
The Mistake: Taking ln(0.0787) instead of ln(0.0787×10−5).
Why it's wrong: The given k values are 105×k. So actual k=(table value)×10−5.
✓ How to avoid: Write clearly:
kactual=(table value)×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: Ea is typically in J/mol or kJ/mol. The correct R for energy calculations is:
R=8.314 J mol−1K−1
✓ How to avoid: Remember:
- For Ea in J/mol: use R=8.314
- For Ea in kJ/mol: use R=0.008314
5. ✗ Confusing slope sign when finding Ea
The Mistake: Taking Ea=slope×R instead of Ea=−slope×R.
Why it's wrong: From lnk=lnA−REa⋅T1, the slope is negative:
slope=−REa
So Ea=−slope×R (which gives a positive value).
✓ How to avoid:
- Plot the graph
- Calculate slope =Δ(1/T)Δlnk (will be negative)
- Then Ea=−slope×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 lnk vs 1/T. …
Showing the 12 most recent of 13 on this concept.
- AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQQ.In the Arrhenius equation, exp(−RTEa), is equal to (A) Frequency factor (B) Fraction of molecules that have energy higher than Ea (C) Rate of the reaction (D) Rate constant of the reaction
›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=Ae−Ea/RT, splits the rate constant into two physically distinct factors: A (the frequency/pre-exponential factor, related to collision frequency and orientation) and e−Ea/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 ≥Ea 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=Ae−Ea/RT.
- Recognize A is the frequency factor — related to the rate of collisions and their proper orientation, independent of energy considerations.
- Recognize e−Ea/RT arises from integrating the Maxwell-Boltzmann distribution above the energy threshold Ea — it represents the fraction of molecules whose energy exceeds Ea at temperature T. …
- AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQQ.At 300°C, decomposition of azomethane follows first order kinetics. Rate constant of this reaction at this temperature is 2.5×10−4s−1. If the activation energy of the reaction is 42 kcal mol−1, what is the temperature (in K) at which the half-life of the reaction is 138.6 seconds? (R=2 cal K−1mol−1, log20=1.30) (A) 725 (B) 425 (C) 525 (D) 625
›Reveal solutionSolution
This tests the Arrhenius equation relating rate constants at two temperatures; solving gives T2≈625K.
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 Ea needs a bigger temperature jump to boost k by the same factor.
Step-by-Step Solution
- T1=300°C=573K, k1=2.5×10−4s−1.
- Desired half-life t1/2=138.6s for a first-order reaction: k2=138.60.693=5×10−3s−1.
- Ratio: k2/k1=2.5×10−45×10−3=20.
- Arrhenius (two-temperature form): logk1k2=2.303REa(T11−T21).
- 2.303REa=2.303×242000=9118.5. …
- AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQQ.The rate constant of a first order reaction at 600 K is 1.6×10−5s−1. If its activation energy is 198.87 kJ mol−1, what is the rate constant (in s−1) at 700 K? (R=8.3 J mol−1K−1) (antilog(0.4771) = 3.0) (A) 4.8×10−4 (B) 4.8×10−3 (C) 4.8×10−2 (D) 3.2×10−4
›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 k2=4.8×10−3 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 Ea 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(k2/k1)=2.4771.
Step-by-Step Solution
- Write the two-temperature Arrhenius relation: logk1k2=2.303REa(T11−T21).
- Compute T11−T21=6001−7001=600×700700−600=420000100=42001.
- Compute 2.303REa=2.303×8.3198870=19.1149198870≈10403.9.
- Multiply: 10403.9×42001≈2.4771. …
- AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQQ.The k value of the reaction A → products is 5×103 s−1 at 300 K. Its activation energy is 50 kJmol−1. At T(K), its value becomes 1.0×104 s−1. What is the value of T (in K)? (R=8.3 Jmol−1K−1), (log2=0.3) (A) 397 (B) 311 (C) 286 (D) 345
›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=Ae−Ea/RT can be converted, for two temperatures, into a linear relation between log(k2/k1) and (1/T1−1/T2), letting us solve for an unknown temperature when the rate constants and activation energy are known.
Step-by-Step Solution
- Two-point Arrhenius equation: logk1k2=2.303REa(T11−T21).
- Given k1=5×103 s−1 at T1=300 K, and k2=1.0×104 s−1 at T2=T (to find). Ratio k2/k1=5×1031.0×104=2.
- Given log2=0.3, so LHS =0.3.
- Compute 2.303REa=2.303×8.350000=19.11550000≈2615.7.
- So 0.3=2615.7(3001−T1), giving 3001−T1=2615.70.3≈1.1468×10−4. …
- AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQQ.At 600 K, the time taken for the completion of 10% of a first order reaction is same as that of its 20% completion at 610 K. What is the value of ratio of rate constants (k600k610)? (log(1.111)=0.0457; log(1.25)=0.0969) (A) 1.211 (B) 2.118 (C) 2.511 (D) 2.711
›Reveal solutionSolution
Same reaction time for 10% completion at 600K and 20% completion at 610K lets the time cancel, leaving k610/k600=log(1.25)/log(1.111)≈2.12.
Concept and Intuition
For a first-order reaction, k=t1ln1−x1 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: k600=t1ln0.901=t2.303log(1.111).
- At 610 K, 20% complete in the same time t: k610=t1ln0.801=t2.303log(1.25).
- Ratio: k600k610=log(1.111)log(1.25)=0.04570.0969≈2.12. …
- AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQQ.Given below are two statements Statement I: Order of a reaction can be obtained from experiment and can have zero or positive integer or positive fraction values Statement II: In the Arrhenius equation, the frequency factor is the fraction of molecules that can have energy higher than Ea Correct answer is (A) Statements I and II both are correct (B) Statements I and II both are not correct (C) Statement I is correct but statement II is not correct (D) Statement I is not correct but statement II is correct
›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 Ea; that fraction is given by the exponential term e−Ea/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=Ae−Ea/RT
Here, k is the rate constant, A is the frequency factor (also called the pre-exponential factor), Ea 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−Ea/RT is the fraction of molecules that have energy equal to or greater than Ea. 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 H2+Br2→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 Ea"
- Let’s revisit the Arrhenius equation: k=Ae−Ea/RT. …
- AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQQ.The rate constant of a first order reaction at 400 and 500 K is respectively 2×10−5 s−1 and 4×10−3 s−1. What is the approximate activation energy (in kJmol−1)? (R=8.3 Jmol−1K−1; log2=0.3) (A) 880 (B) 88 (C) 38.2 (D) 8.8
›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: logk1k2=2.303REa(T11−T21).
- k1k2=2×10−54×10−3=200; log200=log2+log100=0.3+2=2.3.
- T11−T21=4001−5001=400×500500−400=200000100=5×10−4 K−1.
- 2.3=2.303×8.3Ea×5×10−4.
- 2.303×8.3=19.115. …
- AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQQ.The rate constant of a first order reaction is 3.46×10−2 s−1 at 298K. What is the rate constant of the reaction at 350 K if its activation energy is 50.1 kJ mol−1? (R=8.314 J K−1mol−1) (log 2 = 0.3010) (A) 0.592 s−1 (B) 0.692 s−1 (C) 0.792 s−1 (D) 0.892 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−Ea/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: logk1k2=2.303REa(T11−T21).
- Compute T11−T21=2981−3501=298×350350−298=10430052≈4.986×10−4 K−1.
- Compute 2.303REa=2.303×8.31450100=19.14750100≈2616.6.
- Multiply: logk1k2≈2616.6×4.986×10−4≈1.305.
- So k1k2=101.305=10×100.305. Using log2=0.3010, 100.305 is just slightly above 2, i.e. ≈2.02, giving k2/k1≈20.2. …
- AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQQ.The rate constant of a first order reaction was doubled when the temperature was increased from 300 to 310 K. What is its approximate activation energy (in kJmol−1)? (R=8.3 Jmol−1K−1; log2=0.3) (A) 5.33 (B) 533.3 (C) 53333 (D) 53.33
›Reveal solutionSolution
Plugging the rate-doubling data into the two-point Arrhenius equation gives Ea≈53.33 kJmol−1 — option (D).
Concept and Intuition
The Arrhenius equation, k=Ae−Ea/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 Ea 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: logk1k2=2.303REa(T11−T21).
- Here k2/k1=2 (rate doubled), so log2=0.3 (given).
- T11−T21=3001−3101=300×310310−300=9300010=1.0753×10−4 K−1.
- 0.3=2.303×8.3Ea×1.0753×10−4.
- 2.303×8.3=19.115. …
- AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQQ.The rate constants of a first order reaction at 300 K is k1 and 400 K is k2. What is the value of lnk1k2 if activation energy of reaction is 41.5 kJmol−1? (R=8.3 JK−1mol−1) (A) 1.809 (B) 4.166 (C) 2.083 (D) 3.618
›Reveal solutionSolution
Plugging Ea, R, T1=300 K and T2=400 K into the two-point Arrhenius equation gives ln(k2/k1)≈4.166.
Concept and Intuition
The Arrhenius equation k=Ae−Ea/RT shows that rate constants at two temperatures are related by lnk1k2=REa(T11−T21), since the pre-exponential factor A cancels out. A higher activation energy or a larger temperature gap gives a bigger ratio k2/k1.
Step-by-Step Solution
- Convert Ea to joules: 41.5 kJ/mol=41500 J/mol.
- Compute REa=8.341500=5000.
- Compute T11−T21=3001−4001=12004−3=12001. …
- AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQQ.The rate constant, k for a first order reaction, C2H5I(g)→C2H4(g)+HI(g) is x s−1 at 600 K and 4x s−1 at 700 K. The energy of activation of the reaction (in kJ mol−1) is (log 4 = 0.6, R = 8.3 J K−1 mol−1) (A) 48.16 (B) 58.16 (C) 38.16 (D) 28.16
›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: lnk1k2=REa(T11−T21). Given rate constants at two temperatures, we can solve directly for Ea.
Step-by-Step Solution
- k1=x at T1=600 K, k2=4x at T2=700 K, so k2/k1=4.
- ln4=2.303log4=2.303×0.6=1.3818.
- T11−T21=6001−7001=600×700700−600=420000100=2.381×10−4 K−1. …
- AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQQ.The rate constant of a reaction at 500K and 700K are 0.02 s−1 and 0.2 s−1 respectively. The activation energy of the reaction (in kJ mol−1) is (R=8.3 J K−1 mol−1) (A) 66.90 (B) 33.45 (C) 22.30 (D) 44.45
›Reveal solutionSolution
Applying the two-point Arrhenius equation gives activation energy ≈33.45 kJ/mol.
Concept and Intuition
The Arrhenius equation k=Ae−Ea/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 Ea.
Step-by-Step Solution
- Write lnk1k2=REa(T11−T21).
- k1k2=0.020.2=10, so ln10=2.303.
- T11−T21=5001−7001=500×700700−500=350000200=5.714×10−4 K−1. …
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