Q.Which of the following statements are in accordance with the Arrhenius equation? (Two or more than two options may be correct.)

The key idea is the Arrhenius equation: $k = A e^{-E_a / RT}$, which shows how the rate constant $k$ depends on temperature $T$ and activation energy $E_a$.

Reasoning:

1. As $T$ increases, the exponent $-\frac{E_a}{RT}$ becomes less negative, so $e^{-E_a/RT}$ increases — hence $k$ increases. This matches statement (i). …

The Arrhenius equation is the backbone of how we understand reaction rates in chemical kinetics. It tells us that the rate constant $k$ depends on two key factors: temperature and activation energy. The equation is:

where $A$ is the pre-exponential factor (frequency of collisions with proper orientation), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature.

The exponential term $e^{-E_a/(RT)}$ is the fraction of molecules that have enough energy to overcome the activation barrier. When you change $T$ or $E_a$, this fraction changes — and that’s what we need to track.

Let’s examine each statement one by one.

1. Statement (i): Rate of a reaction increases with increase in temperature.

   Look at the exponent: $-E_a/(RT)$. As $T$ increases, the denominator $RT$ gets larger, so the whole fraction $E_a/(RT)$ becomes smaller. A smaller negative exponent means $e^{-E_a/(RT)}$ becomes larger (since $e^{-x}$ increases as $x$ decreases). So $k$ increases. Since rate $\propto k$, the rate increases.

   Result: (i) is correct.

2. Statement (ii): Rate of a reaction increases with decrease in activation energy.

   Now hold $T$ fixed. If $E_a$ decreases, the fraction $E_a/(RT)$ becomes smaller, so $e^{-E_a/(RT)}$ becomes larger. Again $k$ increases, so rate increases.

   Result: (ii) is correct.

3. Statement (iii): Rate constant decreases exponentially with increase in temperature.

   This is the opposite of what we just saw. As $T$ increases, $k$ increases — not decreases. The phrase “decreases exponentially” is wrong. In fact, $k$ increases exponentially (in the sense that the Arrhenius plot of $\ln k$ vs $1/T$ is a straight line with negative slope). …

Method: Qualitative Analysis Using the Arrhenius Equation

The Arrhenius equation is:

Where:

• $k$ = rate constant
• $A$ = pre-exponential factor (frequency factor)
• $E_a$ = activation energy
• $R$ = gas constant
• $T$ = absolute temperature

Step 1: Understand the exponential relationship

The term $e^{-E_a / RT}$ tells us:

• As $T$ increases, the exponent $-\frac{E_a}{RT}$ becomes less negative, so $e^{-E_a/RT}$ increases → $k$ increases.
• As $E_a$ decreases, the exponent becomes less negative, so $e^{-E_a/RT}$ increases → $k$ increases.

Step 2: Evaluate each statement

1. Rate of a reaction increases with increase in temperature. ✓ Correct — $T \uparrow \Rightarrow k \uparrow \Rightarrow$ rate $\uparrow$
2. Rate of a reaction increases with decrease in activation energy. …

Common Mistakes on Arrhenius Equation Plot Questions

Mistake 1: Confusing "rate constant" with "rate of reaction"

• The error: Students treat statements about $k$ (rate constant) and actual reaction rate as identical.
• Why it's wrong: Rate depends on $k$ and concentration. The Arrhenius equation directly describes $k$, not the overall rate.
• How to avoid: Always check if the statement says "rate constant" or "rate of reaction." For rate, you need to consider concentration effects too.

Mistake 2: Misinterpreting the exponential relationship

• The error: Thinking $k$ decreases exponentially when temperature increases.
• Why it's wrong: The Arrhenius equation is:

As $T$ increases, the exponent $-E_a/RT$ becomes less negative, so $e^{-E_a/RT}$ increases — meaning $k$ increases.

• How to avoid: Remember: higher temperature → larger $k$ (faster reaction). Option (iii) says "decreases exponentially" — that is false.

Mistake 3: Forgetting that activation energy is in the denominator of the exponent

• The error: Assuming lower $E_a$ always means slower reaction.
• Why it's wrong: Lower $E_a$ makes $-E_a/RT$ less negative → $e^{-E_a/RT}$ larger → $k$ larger → faster reaction.
• How to avoid: Write the equation: $k = A e^{-E_a/RT}$. A smaller $E_a$ gives a larger $k$. So option (iv) ("decreases with decrease in activation energy") is false.

Mistake 4: Not reading "two or more may be correct" carefully

• The error: Selecting only one option when multiple are correct. …