Question

calculate the bond enthalpy of hcl, given that bond enthalpies of H2 and Cl2 are 430 KJ/mol and 242 KJ/mol and formation enthalpy for Hcl is -91KJ/mol

Answer 1

Answer:The bond enthalpy of HCl is 427 kJ/mol.

Explanation:

$\Delta H_f=-91 KJ/mol$....Formation of one mole of HCl.

$H_2+Cl_2\rightarrow 2HCl\Delta H_f=2\times (-91) kJ/mol$..(1)

$H_2\rightarrow 2H$

$\Delta H_{H-H}=430 KJ/mol$..(2)

$Cl_2\rightarrow 2Cl$

$\Delta H_{Cl-Cl}=242 KJ/mol$..(3)

$2HCl\rightarrow 2H+2Cl$

$\Delta H_{H-Cl}=?$..(4)

(2)+(3)-(1)=(4)

$430 KJ/mol+242 KJ/mol-(-182 KJ/mol)=854 kJ/mol$

$2HCl\rightarrow 2H+2Cl,\Delta H_{H-Cl}=854 kJ/mol$

For one mole HCl =$\frac{854 kJ/mol}{2}=427 kJ/mol$

The bond enthalpy of HCl is 427 kJ/mol.

Answer 2

Answer:

427kj/mol

Explanation:

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