Question

Calculate the enthalpy change for the following reaction using the given bond energies $\left(\right.kJ/mol\left.\right)$ :-
$C-H=414,H-O=463,H-Cl=431,C-Cl=326,C-O=335$
$\left(CH\right)_{3}OH\left(\right.g\left.\right)+HCl\left(\right.g\left.\right) \rightarrow \left(CH\right)_{3}Cl\left(\right.g\left.\right)+H_{2}O\left(\right.g\left.\right)$

• A. $-23kJ/mol$
• B. $-42kJ/mol$
• C. $-59kJ/mol$
• D. $-511kJ/mol$

Answer 1

Answer: (A)

$ΔH_{rxn .}=BE$ of reactant $-BE$ of product
$=\left[\right.\left(\right.3\times 414+463+335\left.\right)+\left(\right.431\left.\right)\left]\right.-\left[\right.\left(\right.3\times 414+326\left.\right)+\left(\right.2\times 463\left.\right)\left]\right.$
$=-23KJ/mol$