Final answer:
To find the enthalpy change of the reaction between NH3 and HCl, the heat absorbed by the solution is calculated first, then the number of moles reacting is determined, and finally, the enthalpy change per mole of reactant is found to be -52.62 kJ/mol.
Step-by-step explanation:
To calculate the enthalpy change (ΔH) of the reaction where ammonium (NH3) is mixed with hydrochloric acid (HCl), we use the formula:
ΔH = -q / n
where:
q is the heat absorbed or released by the solution (in Joules), and n is the number of moles of a reactant.
Firstly, we determine q using the specific heat capacity formula:
q = m × c × ΔT
where:
m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature.
Given that the density of the solutions is 1.00 g/mL, we can find the total mass (m) by adding the volumes of the two solutions:
m = (30.0 mL NH3 + 30.0 mL HCl) × 1.00 g/mL = 60.0 g
The temperature increase (ΔT) is:
ΔT = 33.56°C - 21.00°C = 12.56°C
The specific heat (c) is given as 4.18 J/g°C, so we calculate q:
q = 60.0 g × 4.18 J/g°C × 12.56°C = 3156.912 J or 3.157 kJ
Since 2.0 M NH3 reacts completely with 2.0 M HCl in a 1:1 ratio, the moles of NH3 (and HCl) that react are:
n = Molarity (M) × Volume (L) = 2.0 mol/L × 0.030 L = 0.06 mol
Finally, we can calculate ΔH:
ΔH = -q/n = -3.157 kJ / 0.06 mol = -52.62 kJ/mol
Note that the reaction releases heat, hence the negative sign in ΔH.