-balanced chem equation for Hydrochloric acid and ammonia Reaction 2: Sodium hyd
ID: 1059914 • Letter: #
Question
-balanced chem equation for Hydrochloric acid and ammonia
Reaction 2: Sodium hydroxide + Ammonium chloride
Volume (mL) of 2.0 M NaOH(aq) used
25.15
Initial temperature (°C) of the 2.0 M NaOH(aq)
20.57
Volume (mL) of 2.0 M NH4Cl(aq) used
24.75
Maximum temperature (°C) of the 2.0 M NH4Cl(aq)
22.4496
Write the balanced chemical equation for this reaction.
0.75
Determine the T (°C) of the solution surrounding Reaction 2
0.25
Determine the Energy (J) absorbed (+) or released (-) by the solution.
0.75
Determine the Energy (J) absorbed (+) or released (-) by the calorimeter(qcal).
0.75
Determine the Energy (J) absorbed (+) or released (-) by the reaction(qrxn)
0.75
Enthalpy is always reported with respect to a reactant. Calculate the Enthalpy change (kJ/mole), H, for Reaction 2. Your answer should have the correct sign to reflect whether this is an endothermic or exothermic process.
1
Reaction 2: Sodium hydroxide + Ammonium chloride
Volume (mL) of 2.0 M NaOH(aq) used
25.15
Initial temperature (°C) of the 2.0 M NaOH(aq)
20.57
Volume (mL) of 2.0 M NH4Cl(aq) used
24.75
Maximum temperature (°C) of the 2.0 M NH4Cl(aq)
22.4496
Write the balanced chemical equation for this reaction.
0.75
Determine the T (°C) of the solution surrounding Reaction 2
0.25
Determine the Energy (J) absorbed (+) or released (-) by the solution.
0.75
Determine the Energy (J) absorbed (+) or released (-) by the calorimeter(qcal).
0.75
Determine the Energy (J) absorbed (+) or released (-) by the reaction(qrxn)
0.75
Enthalpy is always reported with respect to a reactant. Calculate the Enthalpy change (kJ/mole), H, for Reaction 2. Your answer should have the correct sign to reflect whether this is an endothermic or exothermic process.
1
Explanation / Answer
Balanced chemical equation for ammonia and HCl
NH3 + HCl ---> NH4Cl
1 mole of NH3 require 1 mole of Hcl for neutralization
From the data
Total volume of solution = 25.15 + 24.75 = 49.9 ml
density of water = 1 g/ml
So, mass of solution = 49.9 g
dT = 22.4496 - 20.57 = 1.8796 oC
Energy evolved during the reaction = -49.9 x 4.184 x 1.8796 = -393.426 J
Enthalpy per mole = -393.426 J/2 x 0.02475 = -7.93 kJ/mol of NH4Cl
the reaction is exothermic, dH is -ve.
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