Use Hess’ Law and ΔH for the first two reactions: 1. NaOH (aq) + HCl (aq)...

Use Hess’ Law and ΔH for the first two reactions: 1. NaOH (aq) + HCl (aq) → H2O (l) + NaCl (aq) 2. NaOH (aq) + NH4Cl (aq) → NH3 + NaCl + H2O (l) to determine ΔH for this reaction: NH3 + HCl → NH4Cl

Expert Solution

ENTHALPY OF REACTION

NaOH (aq) + HCl (aq) → H2O (l) + NaCl (aq)

[1ΔHf(H2O (ℓ)) + 1ΔHf(NaCl (aq))] - [1ΔHf(HCl (aq)) + 1ΔHf(NaOH (aq))]
[1(-285.83) + 1(-407.25)] - [1(-167.15) + 1(-470.09)] = -55.8399999999999 kJ
-55.84 kJ (exothermic)

ENTHALPY OF REACTION

NaOH (aq) + NH4Cl (aq) → NH3 + NaCl + H2O (l)

[1ΔHf(NH3 (g)) + 1ΔHf(H2O (ℓ)) + 1ΔHf(NaCl (aq))] - [1ΔHf(NH4Cl (aq)) + 1ΔHf(NaOH (aq))]
[1(-46.11) + 1(-285.83) + 1(-407.25)] - [1(-299.66) + 1(-470.09)] = 30.5599999999999 kJ
30.56 kJ (endothermic)

HCl + NaOH → NaCl + H2O -55.84 KJ

NaOH (aq) + NH4Cl (aq) → NH3 + NaCl + H2O (l) +30.56 KJ

On reversing the equation

NH3 + NaCl + H2O → NH4Cl + NaOH - 30.56 KJ

Using Hess law

NH3 + HCl → NH4Cl (-55.84-30.56)KJ - 86.40 KJ

queen_honey_blossom answered 2 years ago

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