Find the DeltaH?

Question

What is ΔHneut when 50.0 mL of 1.0 M HCl react with 40.0 mL of 1.0 M NaOH to produce a temperature change of +5.6 ? ​[math] d{soln} = 1.05 g/mL, c{sol} = 4.20 J/g. K

Answer 1

When doing problems of this nature, we generally will use the equation q = mC∆T where..

q = heat = ?

m = mass of material = see below for calculation of mass

C = specific heat = 4.20 J/gº

∆T = change in temperature = 5.6º (note, the scale of Celsius or Kelvin doesn't matter as it is a change in temperature, and a 1º change is a 1º change regardless of which scale you are using)

So, let us first find the mass of the solution. We will use the density of 1.05 g/ml...

50.0 ml + 40.0 ml = 90.0 ml

90.0 ml x 1.05 g/ml = 94.5 g (this does not include the mass of HCl, NaOH etc.)

Now we can solve for q:

q = mC∆T = (94.5 g)(4.20 J/gº)(5.6º)

q = 2223 J = 2.223 kJ = 2.2 kJ (2 sig. figs.)

Now to find ∆H_neut, we will determine how many moles of H₂O are formed in the reaction that we are given. The reason for this is that the change in enthalpy for neutralization (∆H_neut) is defined as the heat associated with production of ONE MOLE of water.

HCl + NaOH ==> NaCl + H₂O ... balanced equation

moles HCl = 50.0 ml x 1 L/1000 ml x 1.0 mol/L = 0.050 moles HCl

moles NaOH = 40.0 ml x 1 L/1000 ml x 1.0 mol/L = 0.040 moles HCl

moles H₂O formed = 0.040 moles H₂O b/c NaOH is in limiting supply

∆H_neut = 2.2 kJ/0.04 moles H₂O = 55 kJ