Calculate the standard enthalpy change for the following reaction at 25 degrees C. HCl(g) + NaOH(s)...

Calculate the standard enthalpy change for the following reaction at 25 degrees C.

HCl(g) + NaOH(s) ---> NaCl(s) + H₂O(l)

Expert Solution

From standard table,

H_f^o of

HCl(g) = -92.3 kJ/mol

NaOH(s) = -425.8 kJ/mol

NaCl(s)= -411.2 kJ/mol

H₂O(l) = -285.8 kJ/mol

Now, H_rxn^o = moles *H_f^o (products) - molesH_f^o (reactants)

H_rxn^o = (moles*H_f^o (NaCl(s)) +(moles *H_f^o (H₂O(l))) -(moles *H_f^o (HCl(g))) -(moles *H_f^o (NaOH(s)))

H_rxn^o = -411.2 kJ/mol -285.8 kJmol + 92.3 kJ/mol + 425.8 kJ/mol = -178.9 kJ/mol

queen_honey_blossom answered 2 years ago

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