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Give steps and enthalpy of formation of NaCl (Sodium chloride). |
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Answer» SOLUTION :Ionization : `Na_((G)) RARR Na_((g))^(+) E , Delta_(i) H = ` 495.8 KJ `mol^(-1)` `Cl_((g)) + e^(-) rarr Cl_((g))^(-) , Delta_(e) H = - 348.7 " kJ" mol^(-1)` Crystallization : ` Na_((g))^(+) + Cl_((g))^(-) rarr NaCl_((s)) , Delta_(L) H = - 788 " kJ " mol^(-1)` Total `Delta_(f) H ("NaCl") = Delta_(i) H + Delta_(e) H + Delta_(L) H ` = (495.8 - 348.7 - 788 ) =- 640.9 kJ `mol^(-1)` |
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