When a student mixes 50 ml of 1M HCL and 50ml of 1M NaOH in a coffee cup calorimeter, the temperature of the resultant solution increases from 21^o C to 25^oC. Assuming that the calorimeter absorbs only a negligable quantity of heat, that the total volume of the solution is 100ml, its density 1.0 g/ml, and that its specific heat is 4.18J/g, calculate:

1) The heat change during mixing,

2)The enthalpy change for the reaction:

HCL(aq) + NaOH(aq) ---- NaCl(aq) + H₂O

a). Here mass of solution m = volume x density = 100 x1 = 100gm

We know that change in heat  q_p = C_pΔT = mC_m ΔT (C_m is specific heat, i.e., heat capacity per gram)

Here, C_m = 4.18J/gmK , and  ΔT = 25-21 = 4⁰C = 4K

Here, heat is evolved so temperature change will be (-ve)

So, q_p = 100 x 4.18 x 4 = 1672J = -1672J as the heat is evolved.

b). Enthalpy change for the reaction,  HCl + NaOH -> NaCl + H₂O

Volume of HCl = volume of NaOH = 50 ml = 0.05L

Here, moles of HCl = moles of NaOH = molarity x volume in L = 1 x 0.05 =0.05 moles

Since, in the reaction , only one mole of (OH^- ions) are produced for each mole of reactants ,

So, we have, for each molar reaction,enthalpy change, ΔH =  q_p

But, for 0.05 mole of reactants,

ΔH =  q_p / [OH^- ] =   -1672/ 0.05[OH^- ]   =- 33.44 KJ