Question

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

AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)

When 50.0 mL of 0.100 M AgNO3 is combined with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from 23.40 ∘C∘ to 24.21∘C.

Calculate Enthalpy chnage for the reaction as written. Use 1.00 g/mL as the density of the solution and Cs=4.18J/(g??C) as the specific heat capacity of the solution.

Answer 1

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

Step-by-step 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

=
`(0.100)/(1000)* 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* 1.00`) g = 100 g

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

=
`(-m_(solution)* C_(solution)* \Delta T_(solution))/(0.00500mol)`

=
`\frac{-100g* 4.18\frac{J}{g.^(0)\textrm{C}}* [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]