Question

Reaction 1: Sodium hydroxide + Hydrochloric acid
Volume (mL) of 2.0 M HCl(aq) used								25.2
Initial temperature (°C) of the 2.0 M HCl(aq)								21.63
Volume (mL) of 2.0 M NaOH(aq) used								24.8
Maximum temperature (°C) of the 2.0 M NaOH(aq)								21.96

NaOH and HCl final temperature: 35.14 degrees C

Determine the Energy (J) absorbed (+) or released (-) by the solution (qsoln).

Determine the Energy (J) absorbed (+) or released (-) by the calorimeter (qcal).

Determine the Energy (J) absorbed (+) or released (-) by the reaction (qrxn)

Calculate the Enthalpy change (kJ/mole), ΔH, for Reaction 1.  

Been stuck on these for hours for the same reaction, please help.

Answer 1

The general reaction between NaOH and HCl is

NaOH + HCl -----------> NaCl + H2O

That means 1 mole of NaOH reacts with 1 mole of HCL

Here concentration of HCl taken = 2.0 M

Volume of HCl used = 25.2 mL

Thus mols of HCl taken = conc * volume

= 2.0 M * 25.2 mL

= 2.0 mol/ 1000 ml *  25.2 mL

= 0.0504‬ mols

Similarly

concentration of NaOH taken = 2.0 M

Volume of NaOH used = 24.8 mL

Thus mols of NaOH taken = conc * volume

= 2.0 M * 24.8 mL

= 2.0 mol/ 1000 ml *  24.8 mL

= 0.0496‬‬ mols

That means here NaOH is present in less amount than HCl . Thus it is the limiting reagant. So the reaction will be

0.0496‬ NaOH + 0.0496‬ HCl -----------> 0.0496‬ NaCl + 0.0496‬ H2O

In this reaction the amount of hear released or absorbed (q) is calculated by the formula-

q = m c ∆T where m = mass of the solution c= specific heat of water = 4.184 J/g°C ∆T = change in temperature

Here m = mass of HCl + mass of NaOH

Now since the final product is water because the product NaCl ldissolves in water, the density of the final solution is taken to be 1g /ml

Thus we can take the density of HCl = density of NaOH = density of solution = 1g/ml

Thus mass of HCl = density * volume = 1g/ml * 25.2 ml = 25.2 g

mass of NaOH = density * volume = 1g/ml * 24.8 ml = 24.8 g

Thus total mass of solution = 25.2 g + 24.8 g = 50 g

The temp change = ∆T = Final temp - initial temp

Here the initial temp of NaOH is taken since after addition of HCl to NaOH, the final temp f the solution increases. Thus here ∆T = Final temp - initial temp

= 35.14 ^oC - 21.96 ^oC

= 13.18‬^oC

Thus the heat absorbed by the solution (since temp increases) is =  

q = m c ∆T = 50 g * 4.184 J/g°C * 13.18‬^oC

= 2757.25 J

= 2.757 kJ

Thus heat released by the reaction will be opposite of amount of heat absorbed by the solution =  - 2.757 kJ

Since here the heat released is -ve, that means the reaction is Exothermic.

Now the Enthalpy change per mole of reaction = - 2.757 kJ / 0.0496‬ mole

= -55.58 kJ/mole