The question is:
The reaction 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) was studied in a coffee cup calorimeter. 100. mL portions of 1.00 M aqueous NaOH and H2SO4, each at 24.0C, were mixed. The maximum temperature achieved was 30.6C. Neglect the heat capacity of the cup and the thermometer, and assume that the solution of products has a density of exactly 1 g/mL and a specific heat capacity of 4.18 J/(g-K).
a. Calculate the heat of reaction, q, in J.
To solve, this is what I did:
q = m * c * [delta]T = ((2*39.9971 g / 1 mol + 98.078 g / 1 mol) * 1.00 mol / 1 L * 1 L / 1000 mL * 100. mL) * (4.18 J / 1 g-K) * (30.6 K – 24.0 K) = 491 J
However, the answer is supposed to be 5.5 x 10 ^ 3 J
Did I do something wrong or is the answer key outdated? It was correct for my other answers though.
The reaction 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) was studied in a coffee cup calorimeter. 100. mL portions of 1.00 M aqueous NaOH and H2SO4, each at 24.0C, were mixed. The maximum temperature achieved was 30.6C. Neglect the heat capacity of the cup and the thermometer, and assume that the solution of products has a density of exactly 1 g/mL and a specific heat capacity of 4.18 J/(g-K).
a. Calculate the heat of reaction, q, in J.
To solve, this is what I did:
q = m * c * [delta]T = ((2*39.9971 g / 1 mol + 98.078 g / 1 mol) * 1.00 mol / 1 L * 1 L / 1000 mL * 100. mL) * (4.18 J / 1 g-K) * (30.6 K – 24.0 K) = 491 J
However, the answer is supposed to be 5.5 x 10 ^ 3 J
Did I do something wrong or is the answer key outdated? It was correct for my other answers though.
-
You have taken 0.1 mole each of NaOH and H2SO4. The 0.1 moles of NaOH will react with 0.05 moles of H2SO4 producing 0.05 moles Na2SO4. The combined mass of the solution is 200g and the temperature rise is 6.6 deg. The heat evolved is 200 x 4.18 x 6.6 = 5.518 kJ so the enthalpy change is - 5.518/0.05 = -110.36 kJ/mol Na2SO4 (the - sign is there as enthalpy is - if heat is evolved)