U5.1 · Reaction RatesPractice · 反应速率

B forms at 0.120 ÷ 40.0 = $0.00300$ M/s. Each A that reacts makes 2 B, so A is consumed at half that rate: $0.00150$ M/s.

Every 3 $\ce{H2}$ consumed make 2 $\ce{NH3}$, so the rate of decrease of $[\ce{H2}]$ : rate of increase of $[\ce{NH3}]$ = 3 : 2 = 1.5. In terms of the rate of reaction, $-\dfrac{1}{3}\dfrac{\Delta[\ce{H2}]}{\Delta t} = \dfrac{1}{2}\dfrac{\Delta[\ce{NH3}]}{\Delta t}$.

The $\ce{CO2}$ escapes from the open flask, so the total mass falls, and how fast it falls tracks the rate. The solution stays colorless, its volume hardly changes, and an open flask stays at atmospheric pressure.

Average rate = −Δ[A] ÷ Δt = (0.655 − 0.439) ÷ (300 − 100) = 0.216 ÷ 200 = $0.00108$ M/s. The minus sign makes the rate positive.

0–100 s: (0.800 − 0.655) ÷ 100 = $0.00145$; 200–300 s: (0.536 − 0.439) ÷ 100 = $9.70 \times 10^{-4}$ M/s. As the reactant is used up there are fewer collisions, so the reaction slows. k stays the same, but the rate also depends on concentration.

X falls from 0.02 M, so it is the reactant, N₂O₅. By the coefficients, every 2 mol of N₂O₅ used up gives 4 mol of NO₂ and 1 mol of O₂: when the reaction is complete, [NO₂] = 2 × 0.02 = 0.04 M (Y) and [O₂] = ½ × 0.02 = 0.010 M (Z). At any moment O₂ forms at 1/4 the rate of NO₂, not 4 times; Y rises fastest precisely because it is NO₂.