Mechanisms of Chemical Change化学变化的机理

According to the Bronsted-Lowry theory, the conjugate base of $\mathrm{HSO_4^-}$ is根据布朗斯特-劳里理论（Bronsted-Lowry theory），$\mathrm{HSO_4^-}$ 的共轭碱是

• (A) $\mathrm{H_2SO_4}$
• (B) $\mathrm{SO_4^{2-}}$
• (C) $\mathrm{HSO_4^-}$
• (D) $\mathrm{H_3O^+}$

At 298 K, an aqueous solution has $\mathrm{pH} = 3.0$. What is the $\mathrm{pOH}$ of this solution?298 K 下，某水溶液的 $\mathrm{pH} = 3.0$。该溶液的 $\mathrm{pOH}$ 是多少？

• (A) 3.0
• (B) 7.0
• (C) 11.0
• (D) 14.0

What is the oxidation state of chromium in the dichromate ion, $\mathrm{Cr_2O_7^{2-}}$?重铬酸根离子 $\mathrm{Cr_2O_7^{2-}}$ 中铬的氧化数是多少？

• (A) $+3$
• (B) $+6$
• (C) $+7$
• (D) $+12$

In the reaction $\mathrm{Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)}$, the oxidizing agent is在反应 $\mathrm{Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)}$ 中，氧化剂是

• (A) $\mathrm{Zn}$
• (B) $\mathrm{Cu^{2+}}$
• (C) $\mathrm{Zn^{2+}}$
• (D) $\mathrm{Cu}$

Equal-concentration ($0.10~\mathrm{mol\,dm^{-3}}$) solutions of $\mathrm{HCl}$ and $\mathrm{CH_3COOH}$ are prepared. Which observation would correctly distinguish the strong acid from the weak acid?配制等浓度（$0.10~\mathrm{mol\,dm^{-3}}$）的 $\mathrm{HCl}$ 与 $\mathrm{CH_3COOH}$ 溶液。下列哪一项观察结果能正确区分强酸与弱酸？

• (A) The $\mathrm{HCl}$ solution has a higher pH than the $\mathrm{CH_3COOH}$ solution.$\mathrm{HCl}$ 溶液的 pH 高于 $\mathrm{CH_3COOH}$ 溶液。
• (B) The $\mathrm{HCl}$ solution has a lower pH and conducts electricity better than the $\mathrm{CH_3COOH}$ solution.$\mathrm{HCl}$ 溶液的 pH 更低，导电性也优于 $\mathrm{CH_3COOH}$ 溶液。
• (C) Both solutions have identical pH because the concentrations are equal.两溶液的 pH 相同，因为浓度相等。
• (D) The $\mathrm{CH_3COOH}$ solution reacts faster with magnesium ribbon because it has the higher $\mathrm{[H^+]}$.$\mathrm{CH_3COOH}$ 溶液与镁条反应更快，因为它的 $\mathrm{[H^+]}$ 更高。

A voltaic (galvanic) cell is built from a $\mathrm{Zn/Zn^{2+}}$ half-cell and a $\mathrm{Cu/Cu^{2+}}$ half-cell, connected by a wire and a salt bridge. Reduction occurs at the electrode called the由 $\mathrm{Zn/Zn^{2+}}$ 半电池与 $\mathrm{Cu/Cu^{2+}}$ 半电池组成原电池（voltaic cell），二者以导线和盐桥相连。还原反应发生在被称为哪种电极上，其极性是

• (A) anode, which is negative.负极，即阳极。
• (B) anode, which is positive.正极，即阳极。
• (C) cathode, which is positive.正极，即阴极。
• (D) cathode, which is negative.负极，即阴极。

In the free-radical substitution of methane with chlorine, the initiation step involves在甲烷与氯气的自由基取代反应中，引发步骤（initiation step）涉及

• (A) heterolytic fission of the $\mathrm{Cl-Cl}$ bond promoted by UV light.紫外光引发 $\mathrm{Cl-Cl}$ 键的异裂（heterolytic fission）。
• (B) homolytic fission of the $\mathrm{Cl-Cl}$ bond promoted by UV light, producing two chlorine radicals.紫外光引发 $\mathrm{Cl-Cl}$ 键的均裂（homolytic fission），生成两个氯自由基。
• (C) homolytic fission of a $\mathrm{C-H}$ bond in methane.甲烷中 $\mathrm{C-H}$ 键的均裂。
• (D) formation of a $\mathrm{C-Cl}$ bond with release of a proton.形成 $\mathrm{C-Cl}$ 键并释放一个质子。

Photochemical chlorination of methane typically yields a mixture of $\mathrm{CH_3Cl}$, $\mathrm{CH_2Cl_2}$, $\mathrm{CHCl_3}$, and $\mathrm{CCl_4}$ rather than pure $\mathrm{CH_3Cl}$. This occurs mainly because甲烷的光化学氯化通常生成 $\mathrm{CH_3Cl}$、$\mathrm{CH_2Cl_2}$、$\mathrm{CHCl_3}$ 与 $\mathrm{CCl_4}$ 的混合物，而非纯 $\mathrm{CH_3Cl}$。这主要是因为

• (A) chlorine radicals can go on to react with product molecules such as $\mathrm{CH_3Cl}$, as well as with unreacted $\mathrm{CH_4}$, giving further substitution.氯自由基不仅能与未反应的 $\mathrm{CH_4}$ 反应，也能继续与 $\mathrm{CH_3Cl}$ 等产物分子反应，从而发生进一步取代。
• (B) the initiation step produces a random assortment of different radicals.引发步骤随机生成多种不同的自由基。
• (C) methane can act as both an oxidizing agent and a reducing agent.甲烷既能作氧化剂，也能作还原剂。
• (D) the $\mathrm{HCl}$ formed inhibits any further reaction.生成的 $\mathrm{HCl}$ 会抑制进一步反应。

Which of the following species is amphiprotic (can act as both a Bronsted-Lowry acid and a Bronsted-Lowry base)?下列哪种物种是两性质子（amphiprotic）物种（既能作布朗斯特-劳里酸，又能作布朗斯特-劳里碱）？

• (A) $\mathrm{CO_3^{2-}}$
• (B) $\mathrm{HCO_3^-}$
• (C) $\mathrm{H_2CO_3}$
• (D) $\mathrm{Na^+}$

The hydrolysis of 2-bromo-2-methylpropane (a tertiary halogenoalkane) in aqueous ethanol is first order overall, with rate $= k\,[\mathrm{substrate}]$, independent of $[\mathrm{OH^-}]$. The hydrolysis of bromoethane (a primary halogenoalkane) is second order overall, with rate $= k\,[\mathrm{substrate}]\,[\mathrm{OH^-}]$. This contrast is best explained because2-溴-2-甲基丙烷（叔卤代烷）在水/乙醇中的水解为总一级反应，rate $= k\,[\mathrm{substrate}]$，与 $[\mathrm{OH^-}]$ 无关；而溴乙烷（伯卤代烷）的水解为总二级反应，rate $= k\,[\mathrm{substrate}]\,[\mathrm{OH^-}]$。造成这种差异的最佳解释是

• (A) the tertiary substrate reacts by $S_N2$ (bimolecular) and the primary substrate reacts by $S_N1$ (unimolecular).叔底物按 $S_N2$（双分子）反应，伯底物按 $S_N1$（单分子）反应。
• (B) the tertiary substrate reacts by $S_N1$, in which the rate-determining step forms a carbocation and does not involve $\mathrm{OH^-}$; the primary substrate reacts by $S_N2$, a single concerted step in which $\mathrm{OH^-}$ attacks the substrate directly.叔底物按 $S_N1$ 反应，其决速步生成碳正离子，不涉及 $\mathrm{OH^-}$；伯底物按 $S_N2$ 反应，是 $\mathrm{OH^-}$ 直接进攻底物的单一协同步骤。
• (C) both substrates react by the same mechanism, but the tertiary substrate is more sterically hindered.两种底物按相同机理反应，只是叔底物空间位阻更大。
• (D) the difference arises purely from a change in leaving-group ability.差异纯粹来自离去基团（leaving group）能力的变化。

Propene, $\mathrm{CH_3CH{=}CH_2}$, undergoes electrophilic addition with $\mathrm{HBr}$. According to Markovnikov's rule, the major product is丙烯 $\mathrm{CH_3CH{=}CH_2}$ 与 $\mathrm{HBr}$ 发生亲电加成。根据马尔科夫尼科夫规则（Markovnikov's rule），主要产物是

• (A) 1-bromopropane, $\mathrm{CH_3CH_2CH_2Br}$.1-溴丙烷，$\mathrm{CH_3CH_2CH_2Br}$。
• (B) 2-bromopropane, $\mathrm{CH_3CHBrCH_3}$.2-溴丙烷，$\mathrm{CH_3CHBrCH_3}$。
• (C) 1,2-dibromopropane, $\mathrm{CH_3CHBrCH_2Br}$.1,2-二溴丙烷，$\mathrm{CH_3CHBrCH_2Br}$。
• (D) propan-1-ol, $\mathrm{CH_3CH_2CH_2OH}$.丙-1-醇，$\mathrm{CH_3CH_2CH_2OH}$。

Given the standard reduction potentials $E^{\ominus}(\mathrm{Cu^{2+}/Cu}) = +0.34~\mathrm{V}$ and $E^{\ominus}(\mathrm{Zn^{2+}/Zn}) = -0.76~\mathrm{V}$, what is $E^{\ominus}_\text{cell}$ for the reaction $\mathrm{Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)}$, and is it spontaneous under standard conditions?已知标准还原电势 $E^{\ominus}(\mathrm{Cu^{2+}/Cu}) = +0.34~\mathrm{V}$，$E^{\ominus}(\mathrm{Zn^{2+}/Zn}) = -0.76~\mathrm{V}$，反应 $\mathrm{Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)}$ 的 $E^{\ominus}_\text{cell}$ 是多少？该反应在标准状态下是否自发？

• (A) Not spontaneous; $E^{\ominus}_\text{cell} = -1.10~\mathrm{V}$不自发；$E^{\ominus}_\text{cell} = -1.10~\mathrm{V}$
• (B) Spontaneous; $E^{\ominus}_\text{cell} = +1.10~\mathrm{V}$自发；$E^{\ominus}_\text{cell} = +1.10~\mathrm{V}$
• (C) Spontaneous; $E^{\ominus}_\text{cell} = +0.42~\mathrm{V}$自发；$E^{\ominus}_\text{cell} = +0.42~\mathrm{V}$
• (D) Not spontaneous; $E^{\ominus}_\text{cell} = -0.42~\mathrm{V}$不自发；$E^{\ominus}_\text{cell} = -0.42~\mathrm{V}$