Calculate the enthalpy change for the following reaction using the given bond energies (.kJ/mol.) :- C-H=414,H-O=463,H-Cl=431,C-Cl=326,C-O=335 (CH)3OH(.g.)+HCl(.g.) arrow (CH)3Cl(.g.)+H2O(.g.)

Question

Calculate the enthalpy change for the following reaction using the given bond energies (.kJ/mol.) :- C-H=414,H-O=463,H-Cl=431,C-Cl=326,C-O=335 (CH)3OH(.g.)+HCl(.g.) arrow (CH)3Cl(.g.)+H2O(.g.) (A) -23kJ/mol (B) -42kJ/mol (C) -59kJ/mol (D) -511kJ/mol

Tardigrade Admin · Accepted Answer

ΔHrxn .=BE of reactant -BE of product =[.(.3× 414+463+335.)+(.431.)].-[.(.3× 414+326.)+(.2× 463.)]. =-23KJ/mol

Q. Calculate the enthalpy change for the following reaction using the given bond energies $(k J / m o l)$ :-
$C - H = 414, H - O = 463, H - Cl = 431, C - Cl = 326, C - O = 335$
$(C H)_{3} O H (g) + H Cl (g) \to (C H)_{3} Cl (g) + H_{2} O (g)$

$- 23 k J / m o l$

$- 42 k J / m o l$

$- 59 k J / m o l$

$- 511 k J / m o l$

$Δ H_{r x n .} = BE$ of reactant $- BE$ of product
$= [(3 \times 414 + 463 + 335) + (431)] - [(3 \times 414 + 326) + (2 \times 463)]$
$= - 23 K J / m o l$

Q. Calculate the enthalpy change for the following reaction using the given bond energies (kJ/mol) :- C−H=414,H−O=463,H−Cl=431,C−Cl=326,C−O=335 (CH)3​OH(g)+HCl(g)→(CH)3​Cl(g)+H2​O(g)

−23kJ/mol

−42kJ/mol

−59kJ/mol

−511kJ/mol