Q.Answer on the basis of the following reaction (species labelled (a)–(d) as printed in the Exemplar):
$\mathrm{\underset{(a)}{HO^-} + \underset{(b)}{CH_3CH(Cl)CH_2CH_3} \rightarrow \underset{(c)}{CH_3CH(OH)CH_2CH_3} + \underset{(d)}{Cl^-}}$
(2-chlorobutane; in the printed diagram the central carbon of

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:

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):

Step 4 — Verify with the general relation.

For any reaction, …

Working

For $\text{A} + 3\text{B} \rightarrow 2\text{C} + \text{D}$, the rate of reaction is

The change in $[\text{A}]$ over $1$ minute ($=60\ \text{s}$):

Since $\text{C}$ has a stoichiometric coefficient of $2$, its rate of formation is twice the rate of reaction: …

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.

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

The negative signs for reactants ensure the rate is positive (since $\frac{d[\text{reactant}]}{dt}$ is negative).

2. Identify the species and coefficients. In the given reaction:

We care about $\mathrm{Br}^-$ (reactant, coefficient 5) and $\mathrm{Br}_2$ (product, coefficient 3).

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

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.

4. Solve for the desired relation. Multiply both sides by 3: $$$$

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.

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

The negative sign indicates disappearance (decrease in concentration), while positive indicates formation.

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

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

Multiply both sides by $-5$:

$\frac{d[X]}{dt} = -\frac{5}{3}\frac{d[Z]}{dt}$ …

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:

$250 + \tfrac{x}{2} = 296 \Rightarrow x = 92\ \text{mm}$ …

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.

1. 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.

2. 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.

3. Option (C): Both $Z$ and $T$ — Since decreasing either one slows the reaction, decreasing both certainly won’t speed it up. This is incorrect. …

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:

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.

Step 3 — Compound the factor.

The effect is multiplicative, not additive — each successive $10^\circ$ step multiplies the rate by 2 again: …