Q.A potential-energy profile (energy on the vertical axis, reaction coordinate on the horizontal axis) is drawn for a reaction. Reading from left to right: the reactants sit at the lowest energy level; the curve rises to a maximum (the activated complex) and then falls to the products, which lie at a HIGHER energy level than the reactants. The drop from the peak down to the product level is marked $E_1$, and the further gap from the product level down to the reactant level is marked $E_2$. Thus the peak lies $E_1 + E_2$ above the reactants and $E_1$ above the products. On this basis, mark the correct option.

Why reactions speed up when you heat them

You already know that most reactions go faster when you raise the temperature. The intuitive reason is simple: molecules move faster, collide more often, and collide harder. But that alone doesn't explain the dramatic jump in rate — a 10 °C rise can double or triple the rate, even though the collision frequency only increases by a few percent. Something else is at work.

The missing piece is that not every collision leads to a reaction. Only collisions with enough energy — above a certain threshold — actually break bonds and form products. That threshold is the activation energy $E_a$. Think of it as a hill the reactants must climb before they can roll down into products. At room temperature, only a tiny fraction of molecules have enough energy to get over that hill. Raise the temperature, and that fraction grows exponentially.

The Arrhenius equation

The Swedish chemist Svante Arrhenius captured this relationship in a single compact formula:

Here:

• $k$ is the rate constant (how fast the reaction proceeds at a given temperature)
• $A$ is the pre-exponential factor (roughly, how often collisions happen with the right orientation)
• $E_a$ is the activation energy (the energy barrier, in J/mol or kJ/mol)
• $R$ is the gas constant (8.314 J/mol·K)
• $T$ is the absolute temperature (in Kelvin)

Why the rate jumps so sharply with temperature

The exponential is the key. Suppose $E_a = 50$ kJ/mol. At 300 K, the fraction is $e^{-50000/(8.314 \times 300)} \approx e^{-20.0} \approx 2 \times 10^{-9}$. At 310 K, it becomes $e^{-50000/(8.314 \times 310)} \approx e^{-19.4} \approx 3.8 \times 10^{-9}$. That's nearly double — even though the temperature rose only 3%. The collision frequency $A$ barely changed, but the exponential term nearly doubled. That's why a small temperature rise can cause a large rate increase.

What $A$ and $E_a$ really mean

$A$ (the pre-exponential factor) accounts for how often molecules collide and whether they're oriented correctly. It depends on the size and shape of the molecules. $E_a$ is the minimum energy needed for a successful collision. A reaction with a high $E_a$ is very sensitive to temperature; one with a low $E_a$ is less sensitive.

The logarithmic form …