- Calculations Reaction 1 Reaction2 Reaction 3 Mass H2o 50g C H2o 4 186 Tige Fhci 50 Ml 50g Dt Final Temp Init 1 (413.52 KiB) Viewed 162 times
heat energy released in three exothermic chemical reactions. This
activity provides a demonstration of Hess' Law using three
reactions: the solubility NaOH in water, the solubility NaOH in HCl
and the reaction of a solution of HCl and a solution of NaOH.
Reaction 1: Solid sodium hydroxide dissolves in water to form an
aqueous solution of ions. NaOH(s)→ Na+(aq) + OH-(aq) + x1 kJ
Reaction 2: Solid sodium hydroxide reacts with an aqueous
solution of hydrogen chloride to form water and an aqueous solution
of sodium chloride. NaOH(s) + H+(aq) + Cl-(aq)→ H2O(l) + Na+(aq) +
Cl-(aq) + x2 kJ
Reaction 3: An aqueous solution of sodium hydroxide reacts with
an aqueous solution of hydrogen chloride to form water and an
aqueous solution of sodium chloride. Na+(aq) + OH-(aq) + H+(aq) +
Cl-(aq)→ H2O(l) + Na+(aq) + Cl-(aq) + x3 kJ . In order to
accurately measure the heat released in each reaction, we will be
using a calorimeter. (For this experiment a styrofoam cup will act
as the calorimeter). The change in temperature that occurs for each
reaction will be used to calculate the energy released in
kilojoules per mole of sodium hydroxide used. We can assume for our
calculations that any heat transferred to the styrofoam and
surrounding air will be negligible. We can also assume that the
specific heat of water is 4.18 J/g°C.
I am having trouble with the question.
Calculations Reaction 1 ✓ Reaction2 Reaction 3 Mass H2O=50g, C H2O= 4.186 Tige fHCI=50 ml = = 50g dt = final temp - initial temp = 30.30- 25 = 5.3 deg C dt- unal temp - initial temp=37-25 = 12° C Volume of NaOH= 25 ml of 1.0M = 25 x 1/1000 mole = 0.025 mole Total volume = 50 ml = 50g dt = final temp - initial temp = 31.7- 25 = 6.7 °C = mass c* At Q=mass* c* At = 50g*4.186 Jigº*5.3°C = 1109.3J Q=mass* c* At = 50*4.186 12 = 2511.6J = 50 x 4.186 x 67J = 1402.3 J Mole of NaOH =mM = 1g 40mol = 0.025 Mole Mole of NaOH =mM = 1g/40mol = 0.025 Mole AH= 1402.3 0.025 J A H=25116.1/0.025 mol = 100464 J = 100.5 KJ AH= 1109.3/0.025 44372J = 44.4 KJ = 56092 J = 56, OSJ Question: Look up Hess' Law and discuss how this experiment is an illustration I