Q.
Bond Enthalpy values of H-H, Cl-Cl are 436kJ/mole and 243kJ/mole respectively. Bond Enthalpy of H-Cl will be ∆f0HCl=-93kJ/mole
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answer is 2.
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Detailed Solution
12H2g+12Cl2g →HClg; ∆H=-93kJ/mole H2g+Cl2g →2HClg; ∆H=-186kJ/mole ∆H=HR-HP -186=HH-H+HCl-Cl-2X -186=+436+243-2X 2X= (436+243+186)kJ/mole X=432.5 kJ/mole
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