Question

Please help on 4

DATA TABLE Trial 3 Trial 1 Trial 2 Maximum temperature ("С)| 29.8°CI Initial temperature (C) 21C 'emperature change (AT) I 궁3°CI 300:1 29.70C 19.8"C T9C 13.20g| IL 8°Cİ I. .30u meles DATA ANALYSIs 1. Write the balanced equation for the reaction of phosphoric acid and sodium hydroxide. 2. Use the equation below to calculate the amount of heat ener 2. Use the cqdation below o Clieulate the amdiny , produced in the reaction. In determining the mass, n, of the solution use 1·11 g/mL for the density. Use 4.18 J the specific heat, Cp, of the solution. g-C) as 3 3. Use the heat energy that you calculated in 2 above to determine the enthalpy change, AH, for the reaction in terms of kJ/mol of phosphoric acid. This is your experimental value of Δ11. 4. Use a table of standard thermodynamic data to calculate the AFH of neutralization for phosphoric acid. Consider this the accepted value of ΔΗ. How does your experimental value for ΔΗ compare to the accepted value? 5. Calculate the percent discrepancy between the calculated (accepted) value of the A/l of neutralization of HjPO4 and your experimental value.

Answer 1

By using standard thermodynamic data, the enthalpy of neutralization of H₃PO₄ can be calculated as follows.

$\Delta$H_{neutralization} = 3*($\Delta$H)_H2O + ($\Delta$H)_Na3PO4 - {($\Delta$H)_H3PO4 + 3*($\Delta$H)_NaOH}

= 3*(-285.8) + (-1987.4) - ((-1277.4) + 3*(-469.4))

= -159.2 kJ/mol

Therefore, for 0.3 mol, the value of $\Delta$H_{neutralizaiton} = -159.2 kJ/mol * 0.3 mol = -47.76 kJ/mol

i.e. The amount of heat released = 47.76 kJ/mol (Theoretical value), 7.89 kJ/mol (experimental value)

i.e. $\Delta$H_accepted (-47.76 kJ/mol) < $\Delta$H_experimental (-7.89 kJ/mol)

Therefore, the theoretical value (accepted value) of the heat released in the neutralization of H₃PO₄ (47.76 kJ/mol) is more than that of experimental value (7.89 kJ/mol).