Given that heat of formation of Ag2O(s), HCl(g) and H2O(l) are −731,−22.06 and −68.32 kcal respectively. Also, Ag2O+2HCl(s)⟶2AgCl(s)+H2O(l);ΔH=−77.61kcal If heat of formation of AgCl in kcal is x then −x/100 is (nearest integer)__________.
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Solution
The expression for the enthalpy change for the reaction is ΔH=2ΔHf(AgCl)+ΔHf(H2O)−[ΔHf(Ag2O)+2ΔHf(HCl)] Substitute values in the above expression. −77.61kcal=2xkcal−68.32kcal−(−731kcal+2(−22.06kcal)) x=−392.20kcal −x/100=3.92≈4kcal