Question

A calormeter has a mass of 200g and a specific heat capacity of 0.420 J g-¹. Into it are put 50.0 cm³ of 1.25 mol dm-³ hydrochloric acid and 50.0cm³ of 1.25 mol dm-³ potassium hydroxide solution at the same temperature. The temperature of the calorimeter and contents rises by 10.0⁰C. Calculate the standard enthalpy of normalization.


Assume the specific heat capacity of all solutions is 4.18 J g-¹ K-¹ and the density of all solutions is 1 g cm-³

Answer 1

$NaOH + HCl =NaCl + H_2O$

$Q_{cal} + Q_{solition} = -q_{rxn}$

$c_{cal}\times m_{cal} + c_{water} \times m_{water }\times \Delta T = -q_{rxn}$

$0.420\times 200 + 4.18 \times (50.0+50.0)\times 10 = -q_{rxn}$

$4264 = -q_{rxn}$

$q_{rxn} = -4264 kJ$

AS moles of NaOH and HCl are the same, then we get the same moles of NaCl

$n(NaCl) = n(NaOH)= n(HCl) = 1.25\times 0.05=0.0625 moles$

$\Delta H_{rxn} = \frac{-4264 kJ}{0.0625 moles} = -68224 \frac{J}{mol} = -68.224 \frac{kJ}{mol}$