Q4. For A + B → products at a fixed temperature: with [A] = [B] = 0.10 mol L−1 the initial rate is 2.0 × 10−3 mol L−1 s−1; doubling [A] alone gives 4.0 × 10−3; doubling [B] alone gives 8.0 × 10−3. Find the rate law and k.
Doubling [A] doubles the rate (first order in A); doubling [B] quadruples it (second order in B). Rate = k[A][B]2, third order overall, and k = 2.0 × 10−3/(0.10 × 0.102) = 2.0 L2 mol−2 s−1 (method of NCERT page 67).
Q5. What are the units of the rate constant of a reaction of order 3/2?
(mol L−1)1−3/2 s−1 = mol−1/2 L1/2 s−1 (rule of NCERT page 69).
Q6. A reaction is first order in A and second order in B. How does its rate change if [B] is doubled? If both are doubled?
Rate = k[A][B]2. Doubling [B] makes it 4 times as fast; doubling both makes it 2 × 4 = 8 times as fast (NCERT pages 67–68).
Q7. What is the molecularity of each step of the iodide-catalysed decomposition of H2O2?
Both steps are bimolecular: H2O2 + I− and H2O2 + IO− (NCERT page 70).
Q8. Assertion: the order of a reaction can be a fraction, but its molecularity cannot. Reason: molecularity counts the species that collide in one elementary step. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: you cannot have half a molecule colliding, while order is whatever experiment finds (NCERT page 70).
Q9. For rate = k[A]3/2[B]−1, what happens to the rate if [B] is doubled?
It is halved, since the rate is proportional to [B]−1 (rate law from NCERT Example 3.3, page 68).
Q10. Why are reactions of molecularity three rare, and molecularity four practically unknown?
The chance of three molecules colliding at the same moment is very small, and of more than three smaller still (NCERT page 70).
Q11. Suppose 2NO + Br2 → 2NOBr goes by a fast equilibrium, NO + Br2 ⇌ NOBr2, then a slow step, NOBr2 + NO → 2NOBr. Predict the rate law and order.
Rate = k2[NOBr2][NO], and [NOBr2] = K[NO][Br2], so rate = k2K[NO]2[Br2]: third order, though the slow step is bimolecular (reasoned).