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化学 高校生

(1)の問題だけがよく分かりません。中和反応後の温度変化を示すところから直線を引いたところが最高温度と言えるのは何故ですか? 放熱しながらも、熱が吸収されてしまった分を考慮してると言う事ですか?

4L リード D 第5章 化学反応とエネルギー 137 ● 応用例題18 反応熱の測定 ►►►119 断熱性の容器に0.50 mol/Lの塩酸100mLをとり、水酸化ナトリウムの固体 2.0g を加えたときの温度曲線を図に示した。 (1) この実験での温度上昇度 AT [K] を, to, t1, tz, ts のう ち必要なものを用いて表せ。 (2) AT = 12.1K, 得られた水溶液の体積を100mL,密度 を1.02g/cm,比熱を4.1J/(g・K) として,この反応 を熱化学方程式で表せ。 H=1.0, 0=16, Na=23 とし, 熱量 [kJ] は整数値とせよ。 (36) HORRO (3) 塩酸と水酸化ナトリウム水溶液の中和熱を 56 kJ/mol として、水酸化ナトリウム (固) の溶解熱を求めよ。 温度[C] t3 100 1000 to 0 指針 (1) NaOH 投入後すぐに放熱が始まったと考え, 中和完了後の温度変化を示す直線を 時間 0 まで延ばし最高温度を求める。 時間 NaOHを加えたとき (1) ts-to (2)発熱量は,1.02g/cm×100cm×4.1J/(g・K)×12.1K = 5060.22J≒5.06kJ 質量 比熱 温度上昇度 塩酸 100 mL 中の HCI は, 0.50 mol/Lx L=0.050 mol CIME 加えた NaOH は, 2.0g -=0.050 mol 40g/mol HCI と NaOH は過不足なく中和している。 1mol 当たりの発熱量は, 5.06kJ -=101.2kJ/mol≒101kJ/mol 311 0.050 mol HClaq + NaOH (固) = NaClaq + H2O (液) + 101kJ 答 (3) (2)の熱量は, 「NaOH の溶解熱 Q [kJ/mol] + 中和熱」 なので, 101kJ/mol=Q [kJ/mol] +56kJ/mol Q=45kJ/mol答の御合 第2編 26

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英語 高校生

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