Answer:
[tex]\mathbf{ \Delta H_{rxn} = -1936 \ kJ/mol}}[/tex]
Explanation:
The equation for the reaction is given as:
[tex]\mathbf{ CH_{4(g)} + 4 F_{2(g)} \to CF_{4(g)} + 4 HF_{(g) }}[/tex]
At standard conditions; the bond energies are as follows;
Bond Bond Energies (kJ/mol)
C-H 413
F-F 155
C-F 485
H-F 567
[tex]\mathbf{\Delta \ H_{rxn} = \sum \Delta H ( reactant) - \sum \Delta H (product)}[/tex]
[tex]\mathbf{\Delta \ H_{rxn} = \sum [\Delta H \ 4( C-H) + \Delta H \ 4(F-F) ]- \sum[ \Delta H \ 4( C-F)+\Delta H \ 4( H-F)] (product)}[/tex]
[tex]\mathbf{\Delta \ H_{rxn} = \sum{ \Delta \ H (4*413) + \Delta \ H (4*155) - \Delta \ H (4(485)) + \Delta H (4(567) }}[/tex]
[tex]\mathbf{ \Delta H_{rxn} = \sum ({ (1652 + 620) - (1940 + 2268)})}[/tex]
[tex]\mathbf{ \Delta H_{rxn} = \sum ({2272- 4208})}[/tex]
[tex]\mathbf{ \Delta H_{rxn} = -1936 \ kJ/mol}}[/tex]
