Determine the molar enthalpy of neutralization(DeltaHneutralization); where, Volume of HCl=100mL, Molarity of HCl=2.00M, Volume of NaOH=100mL,...

Determine the molar enthalpy of neutralization(DeltaHneutralization); where, Volume of HCl=100mL, Molarity of HCl=2.00M, Volume of NaOH=100mL, Molarity of NaOH=2.00M, Average initial temperature of acid and base=22.50 C, Final temperature=35.50 C, Heat capacity of the calorimeter (Cp)= 50 j/K.

Solutions

Expert Solution

Total mass of solution = 100 + 100 = 200g

Q = Total heat produced = Mass of solution x specific heat x (T2 - T1)

= 200 x 1 x (35.5 - 22.5)

= 200x13 = 2600cal

Enthalpy of neutralisation = (Q/V)1000(1/x)

Q = Heat Produced = 2600cal

V =Volume of acid or base = 100ml

x = normality of acid or base = 2N (because normality = molarity for monobasic acid or monoacidic base)

Enthalpy of neutralisation = (2600/100)1000(1/2) = 13000cal = 13.0kCal

Since heat is liberated heat of neuralisation should be negative.

Heat of neutralisation = - 13.0 kcal

queen_honey_blossom answered 1 week ago

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