Question

(a) Define activation energy.

(b) Write the unit of rate constant for

(i) zero order reaction, (ii) second order reaction.

Hint:

Activation energy is the extra energy over the average energy of reactants. The unit of k for order n is (mol per litre) to the power (1 minus n) per second.

Answer 1

Step 1: Understanding the Concept:
Part (a) asks for a definition from the Arrhenius theory. Part (b) asks for the units of the rate constant, which depend on the order of the reaction.

Step 2: Part (a) - Activation energy.
For a reaction to occur, reactant molecules must collide with enough energy. Some molecules reach this energy and some do not.
Activation energy is the minimum extra energy that the reactant molecules must have, above their average energy, so that a collision gives products. It is the difference between the threshold energy and the average energy of the reactants.
\[ E_a = E_{\text{threshold}} - E_{\text{average}} \]

Step 3: Part (b) - General unit of k.
For a reaction of order \( n \), the rate is \( \text{rate} = k[A]^n \). So the unit of \( k \) is the unit of rate divided by the unit of concentration raised to \( n \).
\[ \text{unit of } k = \frac{\mathrm{mol\ L^{-1}\ s^{-1}}}{(\mathrm{mol\ L^{-1}})^n} = \mathrm{(mol\ L^{-1})^{1-n}\ s^{-1}} \]

Step 4: Apply to each order.
Zero order, \( n = 0 \): the unit is \( \mathrm{mol\ L^{-1}\ s^{-1}} \).
Second order, \( n = 2 \): the unit is \( \mathrm{(mol\ L^{-1})^{-1}\ s^{-1} = L\ mol^{-1}\ s^{-1}} \).

Final Answer:
Activation energy is the extra energy needed over the average energy of reactants for a reaction to occur. Units: zero order \( \mathrm{mol\ L^{-1}\ s^{-1}} \); second order \( \mathrm{L\ mol^{-1}\ s^{-1}} \).