Question

100 mL of a 0.300 M solution of AgNO3 reacts with 100 mL of a 0.300 M solution of HCl in a

Answer 1

Answer:

$\Delta _rH=-39,1kJ$

Explanation:

Hello!

In this case, since it is possible to assume that the heat released by the reaction is absorbed by the water in the calorimeter we can write:

$Q_{rxn}=-200mL*\frac{1g}{1mL}*4.184\frac{J}{g\°C}*(23.20\°C-21.80\°C)\\\\Q_{rxn}=-1,171.5J$

Now, since the reaction between silver nitrate and hydrochloric acid is:

$HCl+AgNO_3\rightarrow AgCl+HNO_3$

We can see there is a reacting 1:1 mole ratio, thus, the reacting moles are computed via the molarity of the solutions:

$n=0.300mol/L*0.100L=0.0300mol$

Finally, the enthalpy of reaction is:

$\Delta _rH=\frac{-1,171.5J}{0.0300mol}=-39,050.7J$

And in kJ:

$\Delta _rH=-39,1kJ$

Best regards!