Question

Calculate the bond enthalpy of Cl-Cl bond from the following data

CH4+Cl2 --------> Ch3Cl+HCl

ΔH= -100.3kJ. Also the bond enthalpies of C-H, C-Cl, H-Cl bonds are 413, 326 and 431 kJ mol -1 respectively.

Answer 1

ΔH = nHproducts - mHreactants
-109.3 = (3*413+326+431) - (4*413+x)
-109.3 = (1239+326+431) - (1652+x)
-109.3 = (1996) - (1652+x)
-109.3 = 344 - x
x = 453.3 kj mol^-1

Answer 2

Answer:The bond enthalpy of Cl-Cl bond is 444.3 kJ/mol.

Explanation:

$CH_4+Cl_2\rightarrow CH_3Cl+HCl,\Delta H=-100.3 kJ/mol$

Bond enthalpies of C-H, C-Cl, H-Cl bonds

$\Delta H_{C-H}=413 kJ/mol$

$\Delta H_{C-Cl}=326 kJ/mol$

$\Delta H_{H-Cl}=431 kJ/mol$

$\Delta H_{Cl-Cl}=? kJ/mol$

$\Delta H=[3\times \Delta H_{C-H}+1\times \Delta H_{C-Cl}+1\times \Delta H_{H-Cl}]-[4\times \Delta H_{C-H}+1\times \Delta H_{Cl-Cl}]$

$-100.3=[3\times 413 kJ/mol+326 kJ/mol+431 kJ/mol]-[4\times 413 kJ/mol+1\times \Delta H_{Cl-Cl}]$

$\Delta H_{Cl-Cl}=444.3 kJ/mol$

The bond enthalpy of Cl-Cl bond is 444.3 kJ/mol.