Q4. For a gas reaction at 450 K, P = 0.01, the molar collision coefficient is zAB = 5.0 × 1010 L mol−1 s−1 and Ea = 75 kJ mol−1. Find the rate constant.
Ea/RT = 75 000/(8.314 × 450) = 20.047, and e−20.047 = 1.97 × 10−9. So k = PzABe−Ea/RT = 0.01 × 5.0 × 1010 × 1.97 × 10−9 ≈ 0.98 L mol−1 s−1 (reasoned from NCERT page 84).
Q5. A reaction has Ea = 90 kJ mol−1. When the temperature rises from 350 K to 360 K, by what factor do (a) the collision frequency and (b) the fraction e−Ea/RT rise?
(a) At fixed concentrations, Z goes as √T: √(360/350) = 1.014, a rise of 1.4%. (b) e(Ea/R)(1/350 − 1/360) = e0.859 = 2.36. So the rate rises about 2.4 times, almost all of it from the energy factor (reasoned from NCERT page 83).
Q6. Assertion: not every collision between reactant molecules gives products. Reason: only collisions with at least the threshold energy and the proper orientation are effective. Are both true, and does the reason explain the assertion?
Both are true, and the reason is the correct explanation of the assertion (NCERT page 83).
Q7. In CH3Br + OH− → CH3OH + Br−, why does a collision in which OH− strikes the bromine end give no product?
The new C–O bond can form only if OH− reaches the carbon atom, which carries the small positive charge, from the side away from bromine. Striking the bromine end is an improper orientation: the molecules “simply bounce back and no products are formed” (NCERT page 83, on Fig. 3.12). Strictly, at very high energies other processes can occur (reasoned).
Q8. State one drawback of collision theory, and say how the theory allows for molecules of complex shape.
It treats atoms and molecules as hard spheres and ignores their structure. For complex molecules it adds the steric factor P, which allows for the need for proper orientation (NCERT page 83).