Explain the difference between instantaneous rate of a reaction and average rate of a reaction.

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Q. Explain the difference between instantaneous rate of a reaction and average rate of a reaction.

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Concept understanding — Average Rate Of Reaction

Reaction Rate Stoichiometry – From Intuition to Precision

Imagine you are watching a simple reaction:

$2 \text{NO}_2 \rightarrow 2 \text{NO} + \text{O}_2$

As NO₂ disappears, NO appears twice as fast as O₂ appears. Why? Because the balanced equation says: for every 2 molecules of NO₂ that break apart, you get 2 molecules of NO and 1 molecule of O₂. The numbers in front of the species — the stoichiometric coefficients — tell you the relative speeds at which reactants vanish and products appear.

That is the core idea: reaction rate stoichiometry is the relationship between the rates of change of different species in a chemical reaction, dictated by their coefficients in the balanced equation.

The Intuitive Picture

Think of a factory assembly line. The balanced equation is like a recipe:

2 units of raw material A → 2 units of product B + 1 unit of byproduct C

If the line runs steadily, every time 2 units of A are consumed, 2 units of B are produced and 1 unit of C is produced. So the rate at which A disappears must be twice the rate at which C appears. The rate at which B appears equals the rate at which A disappears (both have coefficient 2).

Note

The stoichiometric coefficients are not speeds themselves — they are scaling factors that connect the speeds of different species.

The Precise Statement

For a general reaction:

aA + bB \rightarrow cC + dD

The rate of reaction (often called the rate of the process, $r$ ) is defined as:

r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Here:

$\frac{d[X]}{dt}$ is the instantaneous rate of change of concentration of species X (in mol L⁻¹ s⁻¹).
The minus sign is used for reactants (their concentration decreases with time).
The plus sign is used for products (their concentration increases with time).
Dividing by the coefficient normalises the rate — so $r$ is the same number for every species.

r = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

This single value $r$ is the intrinsic rate of the reaction, independent of which species you measure.

How to Use It (Step by Step)

Suppose you are given:

$2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$

And you measure that $\frac{d[\text{H}_2]}{dt} = -0.040 \text{ M/s}$ (negative because H₂ is being consumed).

Step 1: Write the rate relation:

r = -\frac{1}{2}\frac{d[\text{H}_2]}{dt} = -\frac{1}{1}\frac{d[\text{O}_2]}{dt} = \frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt}

Step 2: Plug in the known value:

r = -\frac{1}{2}(-0.040) = +0.020 \text{ M/s}

Step 3: Find the rate for O₂:

-\frac{d[\text{O}_2]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{O}_2]}{dt} = -0.020 \text{ M/s}

Step 4: Find the rate for H₂O:

\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{H}_2\text{O}]}{dt} = 2r = 0.040 \text{ M/s}

Tip

A quick check: the coefficients tell you the relative rates. Here, H₂ disappears twice as fast as O₂, and H₂O appears at the same rate as H₂ disappears (both coefficient 2). Always verify your numbers match the coefficient ratios.

--- …

Why this formula?

Average Rate of Reaction — Why the Formula Holds

Let’s build this from the ground up. The goal is to understand why the average rate formula looks the way it does — not just memorise it.

1. What does "rate of reaction" mean physically?

A chemical reaction changes the concentration of reactants (decreasing) and products (increasing) over time.

Rate = how fast this change happens.
If you measure the change over a finite time interval, you get the average rate.

2. The core idea: change per unit time

For any quantity $X$ that changes from $X_1$ to $X_2$ over time $t_1$ to $t_2$ :

\text{Average rate of change of } X = \frac{\Delta X}{\Delta t} = \frac{X_2 - X_1}{t_2 - t_1}

This is just the slope of the straight line connecting the two points on a concentration vs. time graph.

3. Applying this to a reaction

Consider a simple reaction:

A \rightarrow B

Reactant A is consumed: $[A]$ decreases.
Product B is formed: $[B]$ increases.

For reactant A (disappearing):

\text{Average rate} = -\frac{\Delta [A]}{\Delta t}

Why the minus sign?

Because $\Delta [A] = [A]_2 - [A]_1$ is negative (concentration drops). The rate itself must be positive (speed is never negative). So we multiply by $-1$ .

For product B (appearing):

\text{Average rate} = +\frac{\Delta [B]}{\Delta t}

Here $\Delta [B]$ is positive, so no minus sign needed.

4. The general formula for any reaction

For a balanced reaction:

aA + bB \rightarrow cC + dD

The average rate is defined per mole of reaction — so it’s the same number regardless of which species you track.

We divide each $\frac{\Delta [\text{species}]}{\Delta t}$ by its stoichiometric coefficient:

\text{Average rate} = -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = +\frac{1}{c} \frac{\Delta [C]}{\Delta t} = +\frac{1}{d} \frac{\Delta [D]}{\Delta t}

Why divide by the coefficient?

If 2 moles of A disappear for every 1 mole of C formed, then $\frac{\Delta [A]}{\Delta t}$ is twice as large as $\frac{\Delta [C]}{\Delta t}$ . Dividing by the coefficient normalises them to the same "per mole of reaction" rate.

5. Key exam point: the formula in one line

For any species $X$ with stoichiometric coefficient $\nu_X$ (negative for reactants, positive for products):

$\text{Average rate} = \frac{1}{\nu_X} \frac{\Delta [X]}{\Delta t}$ …

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The average rate measures the change in concentration over a finite time interval (a slope of a secant), while the instantaneous rate measures the change at a specific moment (the slope of a tangent). The instantaneous rate is the limit of the average rate as the time interval approaches zero.

The Core Idea: Rate as a Slope

Think of a reaction progressing. The concentration of a reactant falls, or a product rises. If you plot concentration ( $c$ ) against time ( $t$ ), you get a curve. The rate at any point is simply how steep that curve is — the slope.

But "steepness" depends on whether you look at a chunk of time or a single instant.

1. Average Rate — The Secant Slope

The average rate is the change in concentration divided by the change in time over a finite interval, say from $t_1$ to $t_2$ .

\text{Average rate} = -\frac{1}{\nu} \cdot \frac{\Delta [\text{Reactant}]}{\Delta t} \quad \text{or} \quad +\frac{1}{\nu} \cdot \frac{\Delta [\text{Product}]}{\Delta t}

where $\nu$ is the stoichiometric coefficient (to normalise rates for different species).

On the concentration-time graph, this is the slope of the secant line joining the two points $(t_1, c_1)$ and $(t_2, c_2)$ .

Watch out

A common mistake is to think the average rate is constant throughout the interval. It is not — it's just the overall change divided by time. The actual rate may vary wildly inside that interval.

Example: For the reaction $A \to B$ , if $[A]$ drops from $0.50\ \text{M}$ to $0.30\ \text{M}$ in 10 seconds, the average rate over those 10 s is:

\text{Average rate} = -\frac{0.30 - 0.50}{10} = +\frac{0.20}{10} = 0.020\ \text{M s}^{-1}

This tells you the mean speed of the reaction during that period, but not how fast it was going at, say, $t = 2\ \text{s}$ .

2. Instantaneous Rate — The Tangent Slope

The instantaneous rate is the rate at a specific moment in time. It is defined as the limit of the average rate as the time interval shrinks to zero:

\text{Instantaneous rate} = -\frac{1}{\nu} \cdot \frac{d[\text{Reactant}]}{dt} \quad \text{or} \quad +\frac{1}{\nu} \cdot \frac{d[\text{Product}]}{dt}

On the graph, this is the slope of the tangent line to the curve at that exact time.

Tip

To find it experimentally, you draw a tangent to the concentration-time curve at the desired time and calculate its slope. For a reaction that slows down over time (as most do), the instantaneous rate at the start (initial rate) is the steepest. …

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Method: Average Rate of Reaction (Using Stoichiometric Coefficients)

Method Name: Stoichiometric Average Rate Method

Concept-first understanding:

The average rate of a reaction tells us how fast the concentration of a reactant or product changes over a specific time interval. Because different substances in the same reaction change at different rates (due to their stoichiometric coefficients), we define a single, positive average rate for the entire reaction.

Steps to calculate the average rate of reaction:

Identify the time interval

Choose two time points: $t_1$ and $t_2$ (where $t_2 > t_1$ ).
Measure the change in concentration

For any reactant or product, find:

$\Delta [\text{substance}] = [\text{substance}]_{t_2} - [\text{substance}]_{t_1}$
Apply the stoichiometric formula

For a general reaction:

aA + bB \rightarrow cC + dD

The average rate of reaction is:

\text{Average rate} = -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = +\frac{1}{c} \frac{\Delta [C]}{\Delta t} = +\frac{1}{d} \frac{\Delta [D]}{\Delta t}

Use negative sign for reactants (concentration decreases)
Use positive sign for products (concentration increases)
Divide by the stoichiometric coefficient to get the same value for all species

Calculate $\Delta t$

$\Delta t = t_2 - t_1$
Compute the final value

Plug in the numbers. The result has units of concentration per time (e.g., $\text{mol L}^{-1} \text{s}^{-1}$ ).

Example (quick illustration):

For reaction $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$ , if $[\text{H}_2\text{O}_2]$ drops from 0.10 M to 0.05 M in 50 seconds:

\text{Average rate} = -\frac{1}{2} \frac{(0.05 - 0.10)}{50} = -\frac{1}{2} \times \frac{-0.05}{50} = 5 \times 10^{-4} \ \text{M s}^{-1}

Difference Between Instantaneous Rate and Average Rate

| Feature | Average Rate | Instantaneous Rate |

|---------|------------------|------------------------| …

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Here’s a breakdown of the common mistakes students make when explaining the difference between instantaneous rate and average rate of reaction, along with clear strategies to avoid them.

1. Confusing the Time Interval

The Mistake:

Students often say “average rate is for the whole reaction” or “instantaneous rate is at the start.”

They forget that average rate is always over a finite time interval (e.g., $t_1$ to $t_2$ ).
They think instantaneous rate is just “the rate at the beginning” — it’s actually the rate at any single moment.

How to Avoid:

Always define average rate as:

\text{Average rate} = -\frac{\Delta[\text{Reactant}]}{\Delta t} \quad \text{or} \quad +\frac{\Delta[\text{Product}]}{\Delta t}

where $\Delta t$ is a finite, measurable time interval.

Define instantaneous rate as the slope of the tangent to the concentration vs. time curve at a specific time $t$ :

\text{Instantaneous rate} = -\frac{d[\text{Reactant}]}{dt} \quad \text{or} \quad +\frac{d[\text{Product}]}{dt}

Memory trick: “Average = over an interval (two points), Instantaneous = at a point (one tangent).”

2. Forgetting the Sign Convention

The Mistake:

Students write rates as positive for reactants and negative for products, or leave out the sign entirely.

Example: Writing $\frac{\Delta [A]}{\Delta t}$ for a reactant without the negative sign.

How to Avoid:

Rule: Rate is always positive.
- For a reactant $R$ : $\text{Rate} = -\frac{\Delta[R]}{\Delta t}$
- For a product $P$ : $\text{Rate} = +\frac{\Delta[P]}{\Delta t}$
Why? Because $\Delta[R]$ is negative (reactant decreases), the minus sign makes the rate positive.
Exam tip: In definitions, always show the sign explicitly.

3. Mixing Up “Slope of the Curve” vs. “Slope of the Chord”

The Mistake:

Students say “instantaneous rate is the slope of the curve” but then draw a chord (straight line between two points) instead of a tangent.
Or they say “average rate is the slope of the tangent” — wrong.

How to Avoid:

Visualise:
- Average rate = slope of the chord (straight line joining two points on the curve).
- Instantaneous rate = slope of the tangent (line that just touches the curve at one point).
Practice: On a concentration-time graph, physically draw a tangent at $t=0$ and a chord between $t=0$ and $t=10$ s. Compare slopes.

4. Ignoring the Stoichiometric Coefficient

The Mistake:

When a reaction has coefficients (e.g., $2A \rightarrow B$ ), students write the rate as just $-\frac{\Delta[A]}{\Delta t}$ without dividing by the coefficient.

How to Avoid:

For a general reaction $aA + bB \rightarrow cC + dD$ , the rate of reaction is:

\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}

Key point: This applies to both average and instantaneous rates.
Check: If you forget the coefficient, the numerical value of the rate will be wrong — examiners deduct marks.

5. Saying “Instantaneous Rate is Constant” or “Average Rate is Always Different”

The Mistake:

Assuming instantaneous rate never changes (it does — it decreases as reactants are used up).
Thinking average rate is always different from instantaneous rate (they can be equal if the reaction is zero order or if the interval is very small).

How to Avoid: …

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

KCET 2025Set D-41 markMCQ
$E_a$
›Reveal solutionSolution
On an energy profile the forward and backward barriers are both measured to the same peak, so $E_a(\text{reverse}) = E_a(\text{forward}) - \Delta H = 215 - 90 = 125$ kJ.

Step 1 — Read the graph.

From the described energy profile:

The reactants sit at the lower level.

The curve rises to a peak (the activated complex / transition state).

The products sit at a higher level than the reactants — so the reaction is endothermic.

$E_a(\text{forward}) = 215\ \mathrm{kJ}$ — measured from reactants up to the peak.

$\Delta H = +90\ \mathrm{kJ}$ — measured from reactants up to products.

Step 2 — Set an energy scale.

Put the reactants at zero:
$E(\text{reactants}) = 0\ \mathrm{kJ}$ $E(\text{peak}) = 0 + E_a(f) = 215\ \mathrm{kJ}$ $E(\text{products}) = 0 + \Delta H = 90\ \mathrm{kJ}$
Step 3 — The reverse activation energy.

Running the reaction backwards, the molecules start at the products and must climb to the same peak (the transition state is common to both directions — this is the principle of microscopic reversibility):
$E_a(\text{reverse}) = E(\text{peak}) - E(\text{products}) = 215 - 90 = 125\ \mathrm{kJ}$
Step 4 — Verify with the general relation.

For any reaction, …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{A}+3 \mathrm{~B} \rightarrow 2 \mathrm{C}+\mathrm{D}$
›Reveal solutionSolution

Rate of reaction $=4.17\times10^{-5}$ , and rate of formation of $\text{C}$ is twice this $=8.32\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}$ . Official key: (B).

Working

For $\text{A} + 3\text{B} \rightarrow 2\text{C} + \text{D}$ , the rate of reaction is
$\text{Rate} = -\frac{d[\text{A}]}{dt} = +\frac{1}{2}\frac{d[\text{C}]}{dt}$
The change in $[\text{A}]$ over $1$ minute ( $=60\ \text{s}$ ):
$-\frac{\Delta[\text{A}]}{\Delta t} = \frac{0.0150 - 0.0125}{60} = \frac{0.0025}{60} = 4.17\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}$
Since $\text{C}$ has a stoichiometric coefficient of $2$ , its rate of formation is twice the rate of reaction: …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{BrO}_3^{-} \text {(aq) }+5 \mathrm{Br}^{-} \text {(aq) }+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{Br}_2(\mathrm{l})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
›Reveal solutionSolution
The rate of appearance of a product and the rate of disappearance of a reactant are linked by stoichiometric coefficients, with opposite signs. For the given reaction, the correct relation is $\frac{d[\mathrm{Br}_2]}{dt} = -\frac{3}{5} \frac{d[\mathrm{Br}^-]}{dt}$ , which corresponds to option (D).

The key idea is that for a chemical reaction, the rate can be expressed in terms of any reactant or product, but we must account for stoichiometry and sign conventions. The rate of disappearance of a reactant is negative (its concentration decreases), while the rate of appearance of a product is positive (its concentration increases). To relate them, we divide each rate by its stoichiometric coefficient and set them equal in magnitude, then adjust signs.

Let’s work through it step by step.

Write the general rate expression. For a reaction $aA + bB \rightarrow cC + dD$ , the rate is defined as:

$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$
The negative signs for reactants ensure the rate is positive (since $\frac{d[\text{reactant}]}{dt}$ is negative).

Identify the species and coefficients. In the given reaction:

$\mathrm{BrO}_3^- + 5\mathrm{Br}^- + 6\mathrm{H}^+ \rightarrow 3\mathrm{Br}_2 + 3\mathrm{H}_2\mathrm{O}$
We care about $\mathrm{Br}^-$ (reactant, coefficient 5) and $\mathrm{Br}_2$ (product, coefficient 3).

Set up the equality from the general rate. Using the definition:

$-\frac{1}{5}\frac{d[\mathrm{Br}^-]}{dt} = \frac{1}{3}\frac{d[\mathrm{Br}_2]}{dt}$
The left side has a negative sign because $\mathrm{Br}^-$ is a reactant; the right side has no negative sign because $\mathrm{Br}_2$ is a product.

Solve for the desired relation. Multiply both sides by 3:
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COMEDK 2024Set 2024-E1 markMCQ
$5 X+Y \rightarrow 3 Z$
›Reveal solutionSolution
The key idea is that reaction rates for different species are linked by stoichiometric coefficients. For the reaction $5X + Y \rightarrow 3Z$ , the rate of disappearance of X is $\frac{5}{3}$ times the rate of formation of Z, giving $4.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1}$ .

The core concept here is stoichiometric rate relationships. In a chemical reaction, the rate at which reactants disappear and products appear are not independent — they are tied together by the coefficients in the balanced equation. This is because for every 5 molecules of X that react, 3 molecules of Z are formed. So the rate of change of concentration for each species must be scaled by its coefficient to give the same "overall reaction rate."

Why this works:

If we define the rate of reaction as $r = -\frac{1}{5}\frac{d[X]}{dt} = -\frac{1}{1}\frac{d[Y]}{dt} = \frac{1}{3}\frac{d[Z]}{dt}$ , then knowing any one of these rates lets us find the others by simple multiplication or division by the appropriate coefficient ratio.

Write the general rate relationship For the reaction $5X + Y \rightarrow 3Z$ , the rate of reaction $r$ is:

$r = -\frac{1}{5}\frac{d[X]}{dt} = \frac{1}{3}\frac{d[Z]}{dt}$
The negative sign indicates disappearance (decrease in concentration), while positive indicates formation.

Identify what we know We are given the rate of formation of Z:

$\frac{d[Z]}{dt} = 2.4 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1}$

Relate the rate of disappearance of X to the rate of formation of Z From the equality above:

$-\frac{1}{5}\frac{d[X]}{dt} = \frac{1}{3}\frac{d[Z]}{dt}$
Multiply both sides by $-5$ :

$\frac{d[X]}{dt} = -\frac{5}{3}\frac{d[Z]}{dt}$ …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. The equation showing the decomposition of Dinitrogen pentoxide is as follows : N2O5(g) -> N2O4(g) + 1/2 O2(g) If the initial pressure is 250 mm and after half an hour of the reaction the total pressure of the gaseous mixture is 296 mm, what is the average rate of the reaction ? (A) 4.04 * 10^-3 atm min^-1 (B) -2.02 * 10^-3 atm min^-1 (C) 2.19 * 10^-2 atm min^-1 (D) 1.09 * 10^-2 atm min^-1$
›Reveal solutionSolution

The pressure rise fixes the N $_2$ O $_5$ consumed at 92 mm; converting to atm over 30 min gives $4.04\times10^{-3}$ atm min $^{-1}$ .

Pressure bookkeeping ( $\text{N}_2\text{O}_5 \to \text{N}_2\text{O}_4 + \tfrac12\text{O}_2$ ), let $x$ = mm of N $_2$ O $_5$ reacted:
$P_{total} = (250-x) + x + \tfrac{x}{2} = 250 + \tfrac{x}{2}$
$250 + \tfrac{x}{2} = 296 \Rightarrow x = 92\ \text{mm}$ …
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KCET 2020Set A-11 markMCQ
$k = PZe^{-E_a/RT}$
›Reveal solutionSolution
The rate constant $k$ increases when the activation energy $E_a$ is decreased, because $k$ depends exponentially on $-E_a/RT$ ; the correct answer is (D).

The Arrhenius equation $k = A e^{-E_a/RT}$ (where $A = PZ$ in the given notation) tells us exactly how temperature and activation energy control reaction speed. The exponential term $e^{-E_a/RT}$ is the key: it represents the fraction of molecules that have enough energy to overcome the activation barrier. To speed up a reaction, we want $k$ to be larger — so we need that exponential term to be as large as possible.

Since the exponent is negative, making $E_a$ smaller makes the exponent less negative, which increases $e^{-E_a/RT}$ . Making $T$ larger also makes the exponent less negative (because $RT$ is in the denominator), so increasing temperature speeds up the reaction. But the question asks which factor has to be decreased to speed up the reaction.

Let’s check each option carefully.

Option (A): $T$ — Temperature appears in the denominator of the exponent. Decreasing $T$ makes $E_a/RT$ larger, so the exponent becomes more negative, and $k$ gets smaller. That slows the reaction down, not speeds it up. So this is wrong.

Option (B): $Z$ — $Z$ is the collision frequency factor (part of the pre-exponential term $PZ$ ). It multiplies the exponential directly: $k = (PZ) e^{-E_a/RT}$ . Decreasing $Z$ reduces the pre-factor, which reduces $k$ . That also slows the reaction. So this is wrong too.

Option (C): Both $Z$ and $T$ — Since decreasing either one slows the reaction, decreasing both certainly won’t speed it up. This is incorrect. …
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KCET 2018Set A-11 markMCQ
$30^\circ \mathrm{C}$
›Reveal solutionSolution

Rate ratio $= (\text{temperature coefficient})^{\Delta T/10} = 2^{60/10} = 2^{6} = 64$ .

Step 1 — Meaning of the temperature coefficient.

The temperature coefficient $\mu$ of a reaction is defined as the ratio of the rate constants for a $10^\circ$ C rise:
$\mu = \frac{k_{T+10}}{k_{T}}$
For most reactions $\mu \approx 2$ – $3$ (the familiar 'rate roughly doubles for every $10^\circ$ C rise' rule). Here $\mu = 2$ .

Step 2 — Count the $10^\circ$ intervals.
$\Delta T = 90^\circ\mathrm{C} - 30^\circ\mathrm{C} = 60^\circ\mathrm{C}$ $n = \frac{\Delta T}{10} = \frac{60}{10} = 6$
Step 3 — Compound the factor.

The effect is multiplicative, not additive — each successive $10^\circ$ step multiplies the rate by 2 again: …
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Q.Explain the difference between instantaneous rate of a reaction and average rate of a reaction.

The Core Idea: Rate as a Slope

1. Average Rate — The Secant Slope

2. Instantaneous Rate — The Tangent Slope

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