Question

When 40.0 mL of 0.200 M HCl at 21.5°C is added to 40.0 mL of 0.200 M NaOH also at 21.5°C in a coffee-cup calorimeter, the temperature of the resulting solution rises to 22.8°C. Assume that the volumes are additive, the specific heat of the solution is 4.18 Jg -1°C -1 and that the density of the solution is 1.00 g mL -1 Calculate the enthalpy change, ΔH in kJ for the reaction:

Answer 1

Answer:

The enthalpy change for the reaction is ΔH = - 54.3 kJ/mol

Explanation:

The reaction between HCl and NaOH is a neutralization reaction:

$HCl + NaOH \rightarrow NaCl + H2O$

Heat released during neutralization = Heat gained by water

i.e. $\qrxn = -\q(solution)-----(1)$

where:

$\q(solution) = mc\Delta T-----(2)$

m = total mass of solution

$m = density*total\ volume = 1.00g/ml*(40.00+40.00)ml = 80.00\ g$

ΔT = change in temperature = 22.8 - 21.5 = 1.3 C

c = specific heat = 4.18 J/g C

$\q(solution) = 80.00g*4.18J/gC*1.3C = 434.7 J$

As per equation (1): qrxn = -434.7 J

The reaction enthalpy ΔH is the heat released per mole of acid (or base)

$Moles\ of\ HCl = V(HCl) * M(HCl) = 0.040 L*0.200moles/L = 0.008\ moles$

$\Delta Hrxn = \frac{q}{mole}=\frac{-434.7J}{0.008mole}=-54337 J/mol=-54.3 kJ/mol$