A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat capacity (excluding any water) is 480. J/K. The following reaction occurs when the two solutions are mixed.HCl(aq) + NH₃(aq) $\rarr$NH₄Cl(aq) The temperature increase is 2.34^$\circ$C. Calculate $\Delta$H per mole of HCl and NH₃ reacted. A) -154 kJ/mol B) -1.96 kJ/mol C) 1.96 kJ/mol D) 154 kJ/mol E) 485 kJ/mol

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Quizplus · Accepted Answer

The heat absorbed by the calorimeter is q = C * ΔT = 480 J/K * 2.34 K = 1123.2 J. Since 0.0200 moles of HCl and NH3 react, ΔH = -1123.2 J / 0.0200 mol = -56160 J/mol = -56.16 kJ/mol. However, considering the heat absorbed by the solution, the total ΔH is approximately -154 kJ/mol.

A 100 ML Sample of 0 →\rarr→ NH4Cl(aq) the Temperature Increase Is 2

A 100 ML Sample of 0 $\rarr$ NH₄Cl(aq) the Temperature Increase Is 2