Question

In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a coffee-cup calorimeter.
What is the enthalpy change of the precipitation reaction represented above if the final temperature of the mixture is 21.0°C?
(Assume that the total mass of the mixture is 100. g and that the specific heat capacity of the mixture is 4.2 J/(g°C)).

Answer 1

The enthalpy change of the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction is

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To find the enthalpy change we need to apply the following equation

$\Delta H =\frac{Q}{n}$

To find the heat (Q):

$Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J$

Now, to find the number of moles that react (n):

$n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles$

Having these two values we can plug in the first equation:

$\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole$

Have a nice day!