Энтропия и спонтанность
International Baccalaureate · IB Diploma · Chemistry · HL Тема 15 4:36 Английское озвучивание · Английский + китайские субтитры (встроенные)
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A reaction can be endothermic and still be thermodynamically favourable.
一个反应可以是吸热的，但仍然在热力学上是有利的。
Enthalpy alone does not determine the direction favoured at a given temperature.
焓本身并不能决定给定温度下有利的方向。
Entropy is associated with energy dispersal and the number of accessible microscopic arrangements.
熵与能量分散以及可及微观排列的数量有关。
Gibbs energy combines enthalpy, entropy and absolute temperature for a process at constant temperature and pressure.
吉布斯能结合了焓、熵和绝对温度，适用于恒温恒压过程。
Entropy is a state property related to energy dispersal and accessible arrangements.
熵是与能量分散和可及排列相关的状态性质。
Gibbs energy is a thermodynamic quantity combining enthalpy and entropy contributions.
吉布斯能是结合焓和熵贡献的热力学量。
Use Delta G equals Delta H minus T times Delta S with consistent energy units.
使用 ΔG = ΔH − TΔS，注意能量单位一致。
A negative Gibbs energy change indicates thermodynamic favourability for the stated conditions, not a fast rate.
负的吉布斯自由能变化表明在给定条件下热力学上有利，并不意味着反应速率快。
An activation barrier can make a favourable process slow.
活化能垒可能导致有利的过程进行缓慢。
In the lab: State whether values are standard-state quantities and record temperature in kelvin.
在实验室中，请说明数值是否为标准态量，并记录以开尔文为单位的温度。
Convert entropy from joules per kelvin per mole into kilojoules per kelvin per mole when enthalpy is in kilojoules per mole.
当焓变单位为千焦每摩尔时，请将熵从焦耳每开尔文每摩尔转换为千焦每开尔文每摩尔。
Known: Delta H equals plus twenty kilojoules per mole, Delta S equals plus one hundred joules per kelvin per mole, and T equals three hundred kelvins.
已知：ΔH = +20 kJ mol⁻¹，ΔS = +100 J K⁻¹ mol⁻¹，T = 300 K。
Convert Delta S to zero point one zero zero kilojoules per kelvin per mole.
将 ΔS 换算为 0.100 kJ K⁻¹ mol⁻¹。
Delta G equals Delta H minus T times Delta S, which is twenty minus three hundred times zero point one zero zero, equaling negative ten kilojoules per mole.
ΔG = ΔH - TΔS = 20 - 300 × 0.100 = -10 kJ mol⁻¹。
The process is favourable under the stated approximation.
在该近似条件下，该过程是自发的。
Your turn: delta H is 30 kilojoules per mole, delta S is 0.10 kilojoules per kelvin per mole and T is 400 kelvin.
轮到你了：ΔH等于30千焦每摩尔，ΔS等于0.10千焦每开尔文每摩尔，T等于400开尔文。
Find delta G.
求ΔG。
Pause the video and work it out.
请暂停视频，自己算一算。
Check: delta G is 30 minus 400 times 0.10, which is minus 10 kilojoules per mole.
核对：ΔG = 30 − 400 × 0.10 = −10 千焦每摩尔。
Thermodynamic favorability does not establish reaction rate.
热力学有利性不能决定反应速率。
Celsius cannot replace Kelvin in T delta S, and a unit conversion error can change the result by a factor of one thousand.
在T乘以ΔS中，摄氏度不能替代开尔文，单位换算错误可能导致结果相差1000倍。
Not true: A thermodynamically favourable reaction must be fast.
不正确。 热力学上有利的反应不一定快。
Feasibility and speed are different questions.
可行性和速度是两个不同的问题。
Thermodynamics (delta G) says a reaction can proceed; kinetics (activation energy, rate) says how fast.
热力学（delta G）说明反应能否进行；动力学（活化能、速率）说明反应有多快。
Diamond to graphite is favourable yet imperceptibly slow.
金刚石转化为石墨在热力学上是有利的，但速率慢到无法察觉。
Entropy is associated with energy dispersal and the number of accessible microscopic arrangements.
熵与能量分散以及可及的微观排列数相关。
Use Delta G equals Delta H minus T times Delta S, with consistent energy units.
使用 ΔG = ΔH − TΔS 时，请确保能量单位一致。
Watch for: thermodynamic favourability does not establish reaction rate.
注意：热力学有利性并不决定反应速率。
Delta H is plus 40 kilojoules per mole.
ΔH为加40 kJ mol⁻¹。
Delta S is plus 80 joules per kelvin per mole.
ΔS为加80 J K⁻¹ mol⁻¹。
The temperature is 400 kelvin.
温度为400 K。
Find delta G.
求ΔG。
A favourable reaction is not necessarily fast.
一个有利的反应不一定快速。
With delta H at plus 20 kilojoules per mole and delta S at plus 0.100 kilojoules per kelvin per mole, delta G falls as temperature rises.
ΔH为正20千焦每摩尔，ΔS为正0.100千焦每开每摩尔，ΔG随温度升高而降低。
It reaches zero at 200 K and is negative, so favourable, above that.
它在200开时为零，高于此温度时为负，即反应有利。
Assume delta H is plus 24 kilojoules per mole and delta S is plus 80 joules per kelvin per mole, both independent of temperature.
假设 ΔH 为 +24 kJ/mol，ΔS 为 +80 J/(K·mol)，且二者都不随温度变化。
Find the crossover temperature and predict favourability at 250 K and 400 K.
求转折温度，并预测在 250 K 和 400 K 时反应是否自发。
Convert delta S to 0.080 kilojoules per kelvin per mole; the crossover T is 24 divided by 0.080, which is 300 K.
把 ΔS 换算为 0.080 kJ/(K·mol)；转折温度 T 等于 24 除以 0.080，即 300 K。
Delta G is plus 4 kilojoules per mole at 250 K and minus 8 at 400 K.
在 250 K 时 ΔG 为 +4 kJ/mol，在 400 K 时为 −8 kJ/mol。
Thermodynamic favourability does not imply a fast reaction.
热力学上自发并不意味着反应快。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。