N₂O_(5(g)) → 2NO_(2(g)) + 1/2O_(2(g))
In the above first order reaction the initial concentration of N₂O₅ is 2.40 × 10⁻² mol L⁻¹ at 318 K. The concentration of N₂O₅ after 1 hour was 1.60 × 10⁻² mol L⁻¹. The rate constant of the reaction at 318 K is ______________ × 10⁻³ min⁻¹. (Nearest integer)
[Given : log 3 = 0.477, log 5 = 0.699]

Solution

Chemistry ma’am · 4 steps

Use the first-order formula with t in minutes, since the answer is asked per minute.

1. First-order formula. For a first-order reaction, the rate constant comes from the starting and remaining concentrations. Take t = 1 h = 60 min because the answer is in min⁻¹.

   k = 2.303/t log[N₂O₅]₀/[N₂O₅]ₜ

2. Concentration ratio. (2.40 × 10⁻²)/(1.60 × 10⁻²) = 1.5 = 3/2. Use log 2 = 1 - log 5 = 0.301, so log 1.5 = 0.477 - 0.301 = 0.176.
3. Put in numbers. k = (2.303 × 0.176)/60 = 0.4053/60 = 6.76 × 10⁻³ min⁻¹.
4. So. k ≈ 6.76 × 10⁻³ min⁻¹, which rounds to 7.

The common slip: 405. Putting t = 1 hour gives k = 0.405 h⁻¹, that is 405, not 7. Convert 1 hour to 60 minutes first.

The idea to keep. In first-order problems, check which time unit the answer wants before you substitute t. The concentration ratio has no units, so t alone sets the unit of k.

Answer from the NTA final answer key. Solution by PrepWiser.