Question

When 20.0 mL of 1.0 M H₃PO₄ is added to 60.0 mL of 1.0 M NaOH at 25.0 degrees celcius in a calorimeter, the temperature of the aqueous solution increases to 35.0 degrees celcius.

Assuming that the specific heat of the solution is 4.18 J/(g⋅∘C), that its density is 1.00 g/mL, and that the calorimeter itself absorbs a negligible amount of heat, calculate ΔH in kilojoules/mol H₃PO₄ for the reaction.

H₃PO₄(aq)+3NaOH(aq)→3H₂O(l)+Na₃PO₄(aq)

Answer 1

i) Heat absorbed by the solution is equal to heat released by the reaction between H3PO4 and NaOH

ii) Given assumptions are

1) Density and specific heat capacity of solution equal to water

2) Negligible amount of heat is absorbed by water

iii) The heat change of solution is calculated as follows

q= m × ∆T × C

where,

m = mass of solution, 80g ( total volume after mixing the two solution is 80ml and density of the water is 1g/ml ,so, 80g)

∆T = Temperature difference, 35℃ - 25℃ = 10℃

C = specific heat of solution , 4.184J/g ℃

q = 80g × 10℃ × 4.184J/g℃

q = 3347.2J

iv) Heat change(q) = - Enthalphy of reaction of for given amount of reaction(∆H)

So,

∆H = - 3347.2J

v) The balanced reaction between H3PO4 and NaOH is

H3PO4(aq) + 3NaOH ------> 3H2O(l) + Na3PO4(aq)

From the reaction we know that 1 mole of H3PO4 reacts with 3moles of NaOH to form one mole of 1mole of H2O

No of moles of H3PO4 = (1.0mol/1000ml)×20ml = 0.02mol

No of moles of NaOH = ( 1.0mol/1000ml) ×60ml = 0.06ml

So, 0.02 mol of H3PO4 moles of H3PO4 react with 0.06moles of NaOH to form 0.06moles of H2O

vi) ∆H in kilojule/mole of H3PO4 is calculated as follows

∆H = ∆H for 0.02mol of H3PO4/ 0.02mol

∆H = -3347.2J /0.02 mol

∆H = -167.4kJ/mol