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Calculate the enthalpy change for the reaction: H2(g) + Cl2(g) ————-> 2HCl(g). Given that bond energies ofH-H, Cl- Cl and H-Cl bonds are 433, 244 and 431 kj mol-1 respectively.

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ItsItvery simple. Write the reaction. Add bond energies of reactants and bond energy of product multiply by 2 since 2 molecules of HCl are present. In second step, two molecules of HCl are formed. Bond breaking liberates energy so it should be negative. Hence, 433+244+(2×-431) =-185 kj/mol is enthalpy...
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ItsItvery simple.

Write the reaction.

Add bond energies of reactants and bond energy of product multiply by 2 since 2 molecules of HCl are present. In second step, two molecules of HCl are formed. Bond breaking liberates energy so it should be negative.

Hence, 433+244+(2×-431) =-185 kj/mol is enthalpy change of given reaction

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A Nitian who love teaching and believes in visualise physics

433+244+(2×-431) =-185 kj/mol As ΔHreaction=?Hproduct-?Hreaction
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