實驗四反應熱的測定 [計算範例說明]

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海洋大學 106 學年度普通化學實驗-反應熱的測定林佩瑛助教

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實驗四反應熱的測定 [計算範例說明]

Experiment 4 Heat of reactions [Example calculations]

一、普化課本：

Chemistry：The Central Science (Brown, 2013), Chapter 5 Thermochemistry

二、 重要觀念：

1. 能量守恆定律 First Law of Thermodynamics (1) the total energy of the universe is a constant

(2) if the system loses energy, it must be gained by the surroundings, and vice versa (3) q (system) + q (surroundings) = 0

2. 反應熱計算公式 H, is called the enthalpy of reaction, or the heat of reaction.

q = m  Cs  T q ：heat transferred m ：mass

Cs ：specific heat

T ：temperature change (Eqm. temp.- Initial temp.) 3. 有效數字 Significant figures (量筒、溫度計、粗秤天平) 4. 單位 SI unit of energy is the joule (J)

non-SI unit is the calorie (cal): 1 cal = 4.184 J

三、 實驗記錄及計算過程：

★ Density：1.00 g/mL；Specific Heat：1.00 cal/g C (一) 卡計的熱容量 Heat capacity of calorimeter

Test 1 2 3

Volume of cold water 50.0 mL Temp. of cold water 19.2 C Volume of hot water 50.0 mL Temp. of hot water 30.0 C Equilibrium temp. 24.3 C

Heat capacity ???

Heat capacity avg.

Calculation：

q (hot water) + q (cold water) + q (calorimeter) = 0

50.0 mL x 1.00 (g/mL) x 1.00 (cal/g ℃) x (24.3℃-30.0℃) q (hot water) + 50.0 mL x 1.00 (g/mL) x 1.00 (cal/g ℃) x (24.3℃-19.2℃) q (cold water) + Hc x (24.3℃-19.2℃) 卡計與內裝溶液之溫度變化相同 q (calorimeter)

= 0

Hc = 5.88 cal/℃

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(二) Heat of neutralization of HCl/NaOH (中和熱) 文獻值為 -56.8 kJ/mole conc.of HCl (aq) 1.00 M conc.of NaOH (aq) 1.00 M Volume of HCl (aq) 50.0 mL Volume of NaOH (aq) 50.0 mL Initial temp. of HCl (aq) 20.6 C Initial temp. of NaOH (aq) 19.9 C Eqm. temp. after reaction 26.3 C Heat evolved ???

Molar heat of neutralization ???

Calculation：

q (HCl/NaOH) + q (HCl) + q (NaOH) + q (calorimeter) = 0

Heat of neutralization of HCl/NaOH q (HCl/NaOH) + 50.0 mL x 1.00 (g/mL) x 1.00 (cal/g ℃) x (26.3℃-20.6℃) q (HCl)

+ 50.0 mL x 1.00 (g/mL) x 1.00 (cal/g ℃) x (26.3℃-19.9℃) q (NaOH) + 5.88 x (26.3℃-20.6℃) 卡計與內裝溶液之溫度變化相同 q (calorimeter)

= 0

Heat of neutralization of HCl/NaOH = (-285)+ (-320) + (-33.516) = -638.516 cal Limiting reagent：HCl mole = NaOH mole = 1.00 M x 0.0500 L = 0.0500 mole Molar heat of neutralization = -638.516 cal÷0.0500 mole

= -12770.32 cal/mole = -53431.018 J/mole

= -53.4 KJ/mole (Exothermic)

(三) Heat of neutralization of CH3COOH/NaOH (中和熱) 文獻值為 -56.5 kJ/mole Reference(二) 請參照(二)

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(四) Heat of solution of ammonium nitrate (硝酸銨的溶解熱) 文獻值為 +27.2 kJ/mole

Mass of NH4NO3 4.01 g Volume of water 50.0 mL

No. of moles of NH4NO3 4.01/80.0

= 0.0501

Initial temp. of water 21.0 C

Eqm. temp. after reaction 15.3 C Heat evolved ???

Molar heat of solution ???

Calculation：

q (solution) + q (NH4NO3) + q (H2O) + q (calorimeter) = 0

Heat of solution of NH4NO3 q (solution)

+ 0 固體溶解，無溫度變化 q (NH4NO3)

+ 50.0 mL x 1.00 (g/mL) x 1.00 (cal/g ℃) x (15.3℃-21.0℃) q (H2O)

+ 5.88 x (15.3℃-21.0℃) 卡計與內裝溶液之溫度變化相同 q (calorimeter)

= 0

Heat of solution of NH4NO3 = (285) + (33.516) = 318.516 cal Molar heat of solution = 318.516 cal÷0.0501 mole

= 6357.6047 cal/mole = 26600.218 J/mole

= 26.6 KJ/mole (Endothermic)

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(五) Heat of formation of magnesium oxide (氧化鎂生成熱) 文獻值為 -601.5 kJ/mole

★黑斯定律之觀念應用：反應熱具有加成性

Mg (s) + 2H⁺ (aq)  Mg ²⁺(aq) + H2 (g) △H1 MgO (s) + 2H⁺(aq) Mg ²⁺ (aq)+ H2O (l) △H2

Mass of Mg strips 0.20g Mass of MgO 0.70g

No. of moles of Mg 0.20/24.31

= 0.0082

No. of moles of MgO 0.70/40.31

= 0.017 Volume of 1.0 M HCl(aq) 100.0 mL Volume of 1.0 M HCl(aq) 100.0 mL Initial temp. of 1.0 M HCl(aq) 20.6C Initial temp. of 1.0 M HCl(aq) 20.5C

No. of moles of H⁺ 0.100 No. of moles of H⁺ 0.100

Eqm. temp. after reaction 28.9C Eqm. temp. after reaction 25.5C

Heat evolved ??? Heat evolved ???

Molar heat of formation ??? Molar heat of formation ???

Calculation：

q (formation) = △H1(Mg) -△H2 (MgO) + △Hf0 (H2O) Mg (s) + 2H⁺ (aq)  Mg ²⁺(aq) + H2 (g)

- [100.0 mL x 1.00 (g/mL) x 1.00 (cal/g ℃) x (28.9℃-20.6℃) + 5.88 x (28.9℃-20.6℃)] 卡計與內裝溶液之溫度變化相同

△H1(Mg) = q(Mg/HCl)=

-[ q (Mg) + q (HCl) + q (cal)]

MgO (s) + 2H⁺(aq)  Mg ²⁺ (aq) + H2O (l)

- [100.0 mL x 1.00 (g/mL) x 1.00 (cal/g ℃) x (25.5℃-20.5℃) + 5.88 x (25.5℃-20.5℃)] 卡計與內裝溶液之溫度變化相同

△H2(MgO)= q(MgO/HCl)=

-[ q (MgO) + q (HCl) + q (cal)]

Mg ²⁺ (aq) + H2O (l)  MgO (s) + 2H⁺(aq)

-△H2(MgO) = -q(MgO/HCl)=

+[ q (MgO) + q (HCl) + q (cal)]

H2(g) + 1/2O2 (g)  H2O (l) 查課本或文獻知-285.8 kJ/mole △Hf0 (H2O)

Mg (s) + 1/2O2(g)  MgO (s) △Hf0 (MgO)

=△H1 + (-△H2 ) + △Hf0(H2O)

△H1= -879 cal÷0.0082 mole = -107195.12 cal/mole = -448504.39 J/mole =-448.5 KJ/mole

△H2= -529 cal÷0.017 mole = -31117.65 cal/mole = -130196.24 J/mole =-130.2 KJ/mole Molar heat of formation of MgO =△H1 + (-△H2 ) + △Hf0(H2O)

= (-448.5 KJ/mole) + (- -130.2 KJ/mole) + (-285.8 kJ/mole) = - 604.1 KJ/mole (Exothermic)

★ 以上各項實驗誤差率公式 𝐄𝐫𝐫𝐨𝐫 % =^{𝑬𝒙𝒑𝒆𝒓𝒊𝒎𝒆𝒏𝒕}^實驗值^{−𝑹𝒆𝒇𝒆𝒓𝒆𝒏𝒄𝒆}^文獻值

𝑹𝒆𝒇𝒆𝒓𝒆𝒏𝒄𝒆文獻值

實驗四 反應熱的測定 [計算範例說明]