Question

The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:

Answer 1

Answer:

Enthalpy change for the reaction is -67716 J/mol.

Explanation:

Number of moles of $AgNO_{3}$ in 50.0 mL of 0.100 M of $AgNO_{3}$

= Number of moles of HCl in 50.0 mL of 0.100 M of HCl

= $\frac{0.100}{1000}\times 50.0$ moles

= 0.00500 moles

According to balanced equation, 1 mol of $AgNO_{3}$ reacts with 1 mol of HCl to form 1 mol of AgCl.

So, 0.00500 moles of $AgNO_{3}$ react with 0.00500 moles of HCl to form 0.00500 moles of AgCl

Total volume of solution = (50.0+50.0) mL = 100.0 mL

So, mass of solution = ($100.0\times 1.00$) g = 100 g

Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)

= $\frac{-m_{solution}\times C_{solution}\times \Delta T_{solution}}{0.00500mol}$

= $\frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}$

= -67716 J/mol

[m = mass, c = specific heat capacity, $\Delta T$ = change in temperature and negative sign is included as it is an exothermic reaction]