Question

initial temperature = 24.3 degree celcuis

final temperature = 35.7 degree celcuis

50.0 ml of 0.740 M phospheric acid +50.0ml of 2.00M of sodium hydroxide

change in temperature = 11.4 degree celcius

calculate the following

1- what is the mass mixture in thecup ?

2-heat flowof the system in cup in KJ?

3- heat of system in cup (qrxn)in KJ?

4-initial moles of H3PO4 in the cup?

5- initial moles of NaOH in the cup?

6-delta H rxn in KJ(experimental)?

7-delta Hrxn in KJ (theoretical)

8- percent error?

Answer 1

Ans. 1. Assuming the reactants have specific gravity equal to that of water (1 g/ mL), the mass of reaction mixture = 50 g (phosphoric acid solution) + 50 g (NaOH solution)

                        = 100 g

Ans. 2. Heat flow (released) by reaction system

Using, q =         Where, q= heat content                m= mass in gram,

c= specific heat                                 T = temperature

Assuming the solution in cup has specific heat equal to that of water = 4.18 J/ g⁰C

Or, q = 100 g x 4.18 J/ g⁰C x 11.4 ⁰C

            = 4765.2 J = 4.7652 kJ

Ans. 3. Heat gained by cup = heat released by reaction = 4.7652 kJ

Ans. 4. Initial moles of H₃PO₄ = molarity x volume of solution in L

                                                = 0.74 M x 0.050 L

                                                = 0.037 moles

Ans. 5. Initial moles of NaOH = Molarity x volume (in L)

                                    = 2.0 M x 0.050 L

                                    = 0.1 moles

Ans. 6. H₃PO₄ + 3 NaOH ---> Na₃PO₄ + 3 H₂O

According to the balanced reaction, 3 moles NaOH reacts with 1 mol H₃PO₄. Thus, moles of NaOH required to neutralize H₃PO₄ = 3 x moles of H₃PO₄ = 3 x 0.037 moles = 0.111 moles. However, the number of moles of NaOH is only 0.100 moles, thus NaOH is the limiting reagent.

Number of moles of H₃PO₄ required to neutralize 0.1 mol NaOH = (1/3) x 0.1 mol = 0.0333 moles.

Therefore, 0.033333 moles NaOH reacts with 0.033333 moles H₃PO₄.

Since, formation of 1 mol product (Na₃PO₄) during neutralization reaction produces 4.7652 kJ heat.

Amount of heat released during formation of 1 mol Na₃PO₄ = 4.7652 kJ / (0.033333 mol)

                                                                                    = 142.956 kJ mol^-1   

Thus, experimental dH_rxn = -142.956 kJ mol^-1   [-ve sign indicates heat is being released]

Ans. 7.Theoritical dH_rxn = -153.63 kJ mol^-1   

Ans. 8. & error = [(153.63 – 142.956) / 153.63] x 100

                        = 6.94%