Question

TABLE 2 2 4.370 22.2 2 D K 02 9 0 Temperature of hydrochloric acid Temperature of sodium hydroxide Average temperature before reaction Maximum temperature after reaction Temperature change 10000 19 From your measurements recorded above, calculate the enthalpy change (AH) for part 2 kilojoules/mole of limiting reagent. Note: you must determine which of the two is the reagent Assume: 1. The specific heat of the final mixture is the same as that of water. 2. The density of the final mixture is the same as the density of distilled water. 3. The mass of the final mixture is the sum of the masses of the two solutions. NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(lia) Alt: mCXAT

Answer 1

Table 2.

Here, the volume of the HCl and NaOH solutions are not mentioned.

So, let 50 mL of HCl solution is mixed with 50 mL of NaOH solution, then the total volume of the solution is 50+50 = 100 mL

The total mass of the solution (m) = 100 mL * 1 g/mL = 100 g (since density is same as that of water)

The specific heat of the final mixture (C) = 4.184 J/g.^oC (specific heat is the same as that of water)

The change in temperature ($\Delta$T) = 7.276 ^oC

Now, the heat released by the solution (q) = m.C.$\Delta$T = 100 g * 4.184 J/g.^oC * 7.276 ^oC = 3.044 kJ

Since the masses are not given, Let NaOH be the limiting reagent with 2.5 mmol or 2.5*10^-3 mol

i.e. $\Delta$H = -q/n = -3.044 kJ/(2.5*10^-3 mol) = 1217.7 kJ/mol