Zero and First Order Reaction Numericals: 10 Solved Examples with Detailed Solutions
Zero Order Reaction Numericals
Key Formula: [A] = [A]0 – kt
1. Time to reach 0.2 mol/L
Given: [A]0 = 0.6 mol/L, [A] = 0.2 mol/L, k = 0.2 mol L-1 min-1
Solution: t = ([A]0 – [A]) / k = (0.6 – 0.2) / 0.2 = 2 min
2. Time for complete consumption
Given: [A]0 = 0.01 mol/L, k = 4 × 10-4 mol L-1 s-1
Solution: t = [A]0 / k = 0.01 / (4 × 10-4) = 25 s
3. Find rate constant
Given: [A]0 = 1.5 mol/L, [A] = 1.0 mol/L, t = 10 min
Solution: k = ([A]0 – [A]) / t = (1.5 – 1.0) / 10 = 0.05 mol L-1 min-1
4. Time for completion after 75% reacted
Given: 75% reacts in 50 s, so [A] = 25% = 0.25 [A]0
Solution: Time for completion = t for 100% – t for 75% = (1/0.25 – 1) × 50 = 16.67 s
5. Rate constant and time to 0.1 mol/L
Given: [A]0 = 1 mol/L, [A] = 0.2 mol/L, t = 100 s
Solution: k = (1 – 0.2) / 100 = 0.008 mol L-1 s-1
For [A] = 0.1 mol/L: t = (1 – 0.1) / 0.008 = 112.5 s
First Order Reaction Numericals
Key Formula: k = (2.303 / t) * log([A]0 / [A])
6. Find rate constant
Given: [A]0 = 1 mol/L, [A] = 0.25 mol/L, t = 40 min
Solution: k = (2.303 / 40) * log(1 / 0.25) = (2.303 / 40) * log(4) = 0.0576 min-1
7. Half-life calculation
Given: k = 0.231 min-1
Solution: t1/2 = 0.693 / k = 0.693 / 0.231 = 3 min
8. Rate constant for 75% decomposition
Given: 75% decomposed, [A] = 0.25 [A]0, t = 60 s
Solution: k = (2.303 / 60) * log(1 / 0.25) = (2.303 / 60) * 0.602 = 0.0231 s-1
9. Rate constant from 90% completion
Given: [A] = 0.1 [A]0, t = 138.6 s
Solution: k = (2.303 / 138.6) * log(1 / 0.1) = 0.048 s-1
10. Rate constant and half-life from concentration change
Given: [A]0 = 0.8 mol/L, [A] = 0.2 mol/L, t = 20 min
Solution: k = (2.303 / 20) * log(0.8 / 0.2) = 0.115 min-1
t1/2 = 0.693 / 0.115 = 6.03 min
Posted on April 24, 2025 | By more-marks.com Team