1.

Calculate C - Cl bond enthalpy from following reaction : CH_(3) Cl(g) + Cl_(2) (g) to CH_(2) Cl_(2) (g) + HCl(g) Delta H^(@) = - 104 kJ. If C-H,Cl - Cl and H - Cl bond enthalpies are 414, 243 and 431 kJ "mol"^(-1) respectively.

Answer»

SOLUTION :`DELTA H^(@) = sum H_("reactant bond")^(@) - sum H_("product bond")^(@)`
`- 104 = (3 XX H_(c - H)^(@) + H_(C - Cl)^(@) + H_(Cl - Cl)^(@))`
`(2 xx H_(C - H)^(@) + H_(C - Cl)^(@) + H_(H - Cl)^(@))`
`- 104 = (3 xx 414 + X + 243)`
`- (2 xx 414 + 2K + 2K + 431)`
`- 104 = (1242 + X + 243) - (828 + 2X + 431)`
`104 = 226 - X`
`:. X = 330 KJ mol^(-1)`


Discussion

No Comment Found

Related InterviewSolutions