Chemistry Labs

第 1 题

一氧化氮在 820 °C 与氢气反应：2 NO(g)+HX2(g)→NX2O(g)+HX2O(g)\ce{2NO(g) + H2(g) -> N2O(g) + H2O(g)}。在不同初始分压下测得 NX2O\ce{N2O} 生成的初始速率（压强用 torr，时间用秒；不使用浓度）：实验1: pNO=120.0p_{\ce{NO}} = 120.0, pHX2=60.0p_{\ce{H2}} = 60.0 → 速率 =8.66×10−2= 8.66 \times 10^{-2} torr s−1^{-1}；实验2: pNO=60.0p_{\ce{NO}} = 60.0, pHX2=60.0p_{\ce{H2}} = 60.0 → 2.17×10−22.17 \times 10^{-2}；实验3: pNO=60.0p_{\ce{NO}} = 60.0, pHX2=180.0p_{\ce{H2}} = 180.0 → 6.62×10−26.62 \times 10^{-2} torr s−1^{-1}。(a) 求速率方程与速率常数。(b) 求 pNO=200p_{\ce{NO}} = 200 torr、pHX2=100p_{\ce{H2}} = 100 torr 时 NO 消失的初始速率。(c) 求 pNO=800p_{\ce{NO}} = 800 torr、pHX2=1.0p_{\ce{H2}} = 1.0 torr 时 pHX2p_{\ce{H2}} 减半所需时间。(d) 拟议机理为 2 NO⇌NX2OX2\ce{2NO <=> N2O2}（k1,k−1k_1, k_{-1}）随后 NX2OX2+HX2→kX2NX2O+HX2O\ce{N2O2 + H2 ->[k_2] N2O + H2O}。对 NX2OX2\ce{N2O2} 用稳态近似推导速率方程，指出其化为实验速率方程的条件，并用 k1k_1, k−1k_{-1}, k2k_2 表示 kk。
第 4/4 步：稳态推导
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pNX2OX2=k1pNO2k−1+k2pHX2 ⇒ Rate=k1k2 pNO2pHX2k−1+k2pHX2 →k−1≫k2pHX2 k1k2k−1 pNO2pHX2;k=k1k2k−1p_{\ce{N2O2}} = \dfrac{k_1 p_{\ce{NO}}^2}{k_{-1} + k_2 p_{\ce{H2}}}\ \Rightarrow\ \mathrm{Rate} = \dfrac{k_1 k_2\,p_{\ce{NO}}^2 p_{\ce{H2}}}{k_{-1} + k_2 p_{\ce{H2}}}\ \xrightarrow{k_{-1} \gg k_2 p_{\ce{H2}}}\ \dfrac{k_1 k_2}{k_{-1}}\,p_{\ce{NO}}^2 p_{\ce{H2}};\quad k = \dfrac{k_1 k_2}{k_{-1}}
分析

对 NX2OX2\ce{N2O2} 作稳态近似：k1pNO2=(k−1+k2pHX2)pNX2OX2k_1 p_{\ce{NO}}^2 = (k_{-1} + k_2 p_{\ce{H2}})p_{\ce{N2O2}}。当 k−1≫k2pHX2k_{-1} \gg k_2 p_{\ce{H2}}（快速预平衡）时化为观测到的 NO 二级方程，且 k=k1k2/k−1k = k_1 k_2/k_{-1}。