Answer to Question #98598 in General Chemistry for Charley

Question

In a coffee-cup calorimeter, 57.0 mL of 0.100 M AgNO3 and 57.0 mL of 0.100 M HCl are mixed to yield the following reaction.
Ag+(aq) + Cl −(aq) → AgCl(s)
The two solutions were initially at 21.90°C, and the final temperature is 22.70°C. Calculate the heat that accompanies this reaction in kJ/mol of AgCl formed. Assume that the combined solution has a mass of 114.0 g and has a specific heat capacity of 4.18 J/°C · g.

Answer 1

The reaction is expressed with the following equation:

$AgNO_3+HCl \rightarrow AgCl \downarrow+HNO_3;$

Thus, one mole of AgNO₃ and one mole of HCl produce together one mole of AgCl, while in our case the amount of AgNO₃ is:

$n(AgNO_3)=n(AgCl)=C_{AgNO_3}*V_{AgNO_3}=0.100M*0.0570L=5.7*10^{-3}mol;$

The amount in moles of HCl and AgCl precipitate is the same.

Now, we have to calculate the amount of heat released in the cup calorimeter. Hypothetically, all heat was absorbed by the solution:

$Q=m(solution)*c(water)*(t(final)-t(starting))=114g*4.18 \frac{J}{C*g}*(22.70C-21.90C)=381J;$

Finally, the molar heat of formation for AgCl precipitate is:

$H=\frac{Q}{n(AgCl)}=\frac{381J}{5.7*10^{-3}mol}=66.8\frac{kJ}{mol}.$