Measuring enthalpy change
International Baccalaureate · IB Diploma · Chemistry · HL Topic 12 4:06 English narration · English + 中文 subtitles burned in
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A cup warms when two solutions react.
当两种溶液发生反应时，杯子会变热。
The temperature rise measures energy transferred to the surroundings; it does not directly equal the enthalpy change 焓变.
温度升高测量了传递给环境的热量，但它并不直接等于焓变。
Exothermic reactions transfer energy to surroundings.
放热反应将能量传递给环境。
Endothermic reactions take energy from surroundings.
吸热反应从环境中吸收能量。
Bond breaking requires energy, and bond formation releases energy.
断键需要吸收能量，成键则释放能量。
Exothermic means transferring energy to the surroundings.
放热是指向环境传递能量。
Enthalpy change is defined as the heat change at constant pressure for a stated process.
焓变是指在恒定压力下，特定过程的热量变化。
Use energy transferred equals mass times specific heat capacity times temperature change.
使用公式：传递的热量等于质量乘以比热容乘以温度变化。
Convert joules to kilojoules before dividing by reaction amount.
在除以反应物物质的量之前，先将焦耳换算为千焦。
An exothermic molar enthalpy change has a negative sign.
放热的摩尔焓变带有负号。
In the lab: Use insulation and a lid, measure starting temperatures consistently, stir, and record a temperature-time series. Estimate the reaction temperature from an appropriate extrapolation rather than ignoring cooling during measurement.
在实验室中，使用隔热材料和盖子。 一致地测量初始温度。 搅拌混合物。 记录温度-时间序列。 通过适当的外推法估算反应温度，而不是忽略测量过程中的冷却效应。
Known: 100 grams of solution rises by 5.0 degrees Celsius; specific heat capacity is 4.18 joules per gram per degree.
已知：100 g 溶液温度升高 5.0 °C；比热容为 4.18 J/(g·°C)。
q equals m c delta T.
q = mcΔT。
q equals 100 times 4.18 times 5.0, which equals 2,090 joules, or 2.09 kilojoules.
q = 100 × 4.18 × 5.0 = 2,090 J = 2.09 kJ。
If 0.050 moles react, delta H equals negative q divided by n, which equals negative 2.09 divided by 0.050, resulting in negative 41.8 kilojoules per mole.
若反应物为 0.050 mol，则 ΔH = -q/n = -2.09/0.050 = -41.8 kJ/mol。
Your turn.
轮到你了。
50 grams of water rises 4 degrees Celsius.
50 克水升温 4 摄氏度。
Use c equals 4.2 joules per gram per degree to find q.
取比热容 c 等于每克每摄氏度 4.2 焦耳，计算 q。
Heat loss usually lowers the observed temperature rise.
热量损失通常会降低观察到的温度升高幅度。
The solution gaining heat and the reaction losing heat have opposite signs.
吸热的溶液与放热的反应具有相反的符号。
Not true: Bond breaking releases energy.
不正确。 断键释放能量是错误的。
Bond breaking is endothermic - it requires energy.
断键是吸热过程，需要吸收能量。
Bond forming releases energy.
成键则释放能量。
Confusing the two reverses the sign of every energy calculation.
混淆这两者会导致所有能量计算的符号反转。
Exothermic reactions transfer energy to surroundings.
放热反应将能量传递给环境。
Use energy transferred equals mass times specific heat capacity times temperature change.
使用公式：传递的能量等于质量乘以比热容乘以温度变化量。
Watch for: heat loss usually lowers the observed temperature rise.
注意：热量散失通常会降低观测到的温升。
Sixty grams of solution rises by three degrees.
60 g 溶液温度升高 3 度。
Use c equals four point two joules per gram per degree.
使用比热容 c = 4.2 J/(g·°C)。
Zero point zero two zero moles react.
0.020 mol 发生反应。
Find q and the enthalpy change per mole.
求热量 q 和每摩尔的焓变。
Bond breaking requires energy.
断键需要吸收能量。
The result is 756 joules, and minus 37.8 kilojoules per mole.
结果是756焦耳，以及每摩尔负37.8千焦。
The lid and the insulated cup reduce heat loss, so the temperature rise measures the energy released.
盖子和隔热杯减少热量散失，所以温度升高可以衡量释放的能量。
One hundred grams warming by 5.0 kelvin takes 2,090 joules.
100克溶液升温5.0开尔文需要2090焦耳。
Fictional solution gains 1.5 kJ when 0.030 mol reacts.
虚构溶液在0.030摩尔反应时吸收1.5千焦热量。
Find molar reaction enthalpy using the stated heat-transfer model, and explain the sign and effect of uncorrected heat loss.
请使用指定的热传递模型计算摩尔反应焓，并解释未校正热损失对符号及结果的影响。
Delta H equals minus q over n, which is minus 1.5 divided by 0.030, giving minus 50 kilojoules per mole.
ΔH 等于 −q/n，即 −1.5 除以 0.030，得 −50 kJ/mol。
The solution gains heat; the reaction loses it.
溶液吸收热量，反应放出热量。
Uncorrected loss usually underestimates the magnitude.
不校正热损失通常会低估其大小。
Include container heat if relevant.
如有必要，应把容器吸收的热量也计算在内。
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。