Question

For the reaction,
$ H_2F_{2(g)} \to H_{2(g)} +F_{2(g)} ; \Delta E = - 14.2\,K $ is cal/mole at $ 25^{\circ}C $ . The change in enthalpy of the reaction is

• A. $ 13.6 \,kJ/mol $
• B. $ -13.6 \,kJ/mol $
• C. $ 1.36 \,kJ/mol $
• D. $ -56.87 \,kJ/mol $

Answer 1

Answer: (D)

$\Delta E = - 14.2\,Kcal/mol$
$ = - 4.18 \times 14.2\,kJ/mol $
$ = - 59.356\,kJ/mol$
$\because \Delta H = \Delta E + p \Delta V$
or $\Delta H = \Delta E + \Delta _{ng} RT$
$\Delta H = - 59 .356\,kJ /mol + 8.314 \,J/mol \times 298\,K$
$[\therefore \Delta_{ng} = 2 - 1 = 1 ]$
$ = - 59.356\,kJ / mol + 2. 48 \,kJ /mol$
$ = - 56.87\,kJ /mol$