Question

Using bond energies, estimate the enthalpy of reaction for the following chemical reaction. CH4(g) + 4 F2(g) → CF4(g) + 4 HF(g) ΔHrxn = ?

Answer 1

Answer:

$\mathbf{ \Delta H_{rxn} = -1936 \ kJ/mol}}$

Explanation:

The equation for the reaction is given as:

$\mathbf{ CH_{4(g)} + 4 F_{2(g)} \to CF_{4(g)} + 4 HF_{(g) }}$

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

$\mathbf{\Delta \ H_{rxn} = \sum \Delta H ( reactant) - \sum \Delta H (product)}$

$\mathbf{\Delta \ H_{rxn} = \sum [\Delta H \ 4( C-H) + \Delta H \ 4(F-F) ]- \sum[ \Delta H \ 4( C-F)+\Delta H \ 4( H-F)] (product)}$

$\mathbf{\Delta \ H_{rxn} = \sum{ \Delta \ H (4*413) + \Delta \ H (4*155) - \Delta \ H (4(485)) + \Delta H (4(567) }}$

$\mathbf{ \Delta H_{rxn} = \sum ({ (1652 + 620) - (1940 + 2268)})}$

$\mathbf{ \Delta H_{rxn} = \sum ({2272- 4208})}$

$\mathbf{ \Delta H_{rxn} = -1936 \ kJ/mol}}$