Bạn viết thiếu E của C-H
\(\left(1\right)CH\equiv CH+\text{2Cl2 → CHCl2-CHCl2 ΔH = ?}\)
\(\left(2\right)\text{CH≡CH → 2C + 2H}\)______________\(E_{C\equiv C}+2E_{C-H}\)
\(\left(3\right)\text{Cl-Cl → 2Cl }\)___________\(E_{Cl-Cl}\)
\(\left(4\right)\text{CHCl2-CHCl2 → 2C + 2H + 4Cl }\)_______\(4E_{C-Cl}+2E_{C-H}+E_{C-C}\)
Ta thấy: (1) = (2) + 2.(2) - (4)
\(\rightarrow\Delta H=\left(E_{C\equiv C}+2E_{C-H}\right)+2.E_{Cl-Cl}-\left(4E_{C-Cl}+2E_{C-H}+E_{C-C}\right)\)
\(=812+2.E_{C-H}\text{+ 2.242,7 - }\text{(4.339 + 2.E}_{C-H}\text{ + 347) = -405,6 kJ/mol}\)