# 普化問題~反應速率和機構

The recation NO + O3→NO2 + O2 was studuied by performing the flooding technique. When [O3] =1.0*10^14 molecules/cm^3, the change of [NO] with time is shown in Table 1. When [NO]=2.0*10^14 molecules/cm^3, the change of [O3] with time is shown in Table 2.

Table 1

time(ms) [NO] (molecules/cm^3)

0 6.0*10^8

100 5.0*10^8

500 2.4*10^8

700 1.7*10^8

1000 9.9*10^7

Table 2

time(ms) [O3] (molecules/cm^3)

0 1.0*10^10

50 8.4*10^9

100 7.0*10^9

200 4.9*10^9

300 3.4*10^9

(a) Using the graphs to determine the rate order with respect to [NO] and [O3]?

(b) Determine the rate law and evaluate the rate constant for this reaction.

(c) Propose a mechanism and explain how it can match with the experimential data.

### 1 個解答

• ?
Lv 7
1 0 年前
最佳解答

零級反應 A --> B, -d[A]/dt=K[A]0=K, -d[A]=Kt, 兩邊積分, [A]為t的線性函數.

一級反應 A --> B, -d[A]/dt=K[A]1, -d[A]/[A]=Kdt, 兩邊積分, log[A]=-Kt, log[A]為t的線性函數.二級反應 2A --> B, -d[A]/dt=K[A]2, -d[A]/[A]2=Kdt, 兩邊積分,1/[A]=Kt, 1/[A]為t的線性函數.

表一中取log[NO]與時間作圖得一線性關係,斜率為-7.82*10-4.表二中取log[O3]與時間作圖也得一線性關係斜率為-1.56*10-3.由上面的推論知此反應為NO和O3的一級反應.反應速率公式可寫成 d[NO2]/dt=K[NO][O3]

表一中斜率=K[O3], 因[O3]=1.0*1014, K=7.82*10-4/1.0*1014=7.82*10-18 s-1, 表二中斜率=K[NO], 因[NO]=2.0*1014, K=1.56*10-3/2.0*1014=7.8*10-18 s-1, 取平均值, 速率常數 K=7.81*10-18 s-1.

NO + O3 --> NO2 + O2 的反應機制符合實驗結果.