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The standard enthalpy of formation of `H_(2)(g)` and `Cl_(2)(g)` and `HCl(g)` are `218 kJ//mol`, `121.88 kJ//mol` and `-93.31 kJ//mol` respectively. Calculate standard enthalpy change in `kJ` for `(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)toHCl(g)`A. `+431.99B. `-262.14`C. `-431.99`D. `+247.37` |
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Answer» Correct Answer - B `Delta_(r )H=DeltaH_("products")-DeltaH_("reactants")` `=-92.3-[(218)/(2)+(121.68)/(2)]` `=-92.3-[109+60.84]` `=-262.14 kJ//mol` |
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