Question

The enthalpy of neutralization for the reaction of HNO3 with KOH is given below.

HNO3 (aq) + KOH (aq) → H2O (l) + KNO3 (aq) ΔH = -57.0 kJ

102.4 ml of 0.662 M HNO3 is combined with 102.4 ml of 0.662 M KOH in a coffee cup calorimeter. Both of the starting solutions were initially at a temperature of 30.64 °C. The density of each solution is 1.00 g/ml. Calculate the final temperature of the solution in the calorimeter (in °C).

Answer 1

Moles of HNO3 = M x V ( in L) = 0.662 x ( 102.4/1000) = 0.0677888

Moles of KOH also 0.067788

Now dH reaction = -57 KJ per 1 mole

Per 0.0677888 moles    heat = dH x moles = - 57 KJ x 0.0677888 = -3.864 KJ = -3864 J

( -ve sign indicates heat released in reaction)

Now this heat is absorbed by solution

solution volume = 102.4+102.4 = 204.8 ml

mass of solution = volume x density = 204.8 ml x 1g/ml = 204.8 g

Now we have formula

Heat absorbed by solution = specific heat of solution x temperature change x mass of solution

3864 = 4.184 J/gC x ( T-30.64) x 204.8 g

T = 35.15 C

( where T is final temp of solution , Specific heat of solution = water specific heat = 4.184 J/gC)