Question

When 50.0 mL of 0.100 M AgNO3 is mixed with 50.0 mL of 0.100 M HCl in a coffee cup calorimeter, the temperature increases from 23.40 °C to 24.21°C. If the specific heat of the solution is 4.184 J/°C·g and its density is 1.00 g/mL, assuming that the calorimeter undergoes no heat change, calculate the enthalpy change for the following reaction in units of kJ/mol AgNO3:AgNO3(aq) + HCl(aq) à AgCl(s) + HNO3(aq)

Answer 1

Answer:

The enthalpy change during the reaction is -67.78 kJ/mol.

Explanation:

Volume of solution = 50.0 mL + 50.0 mL = 100.0 mL

Density of the solution = 1.00g/mL

Mass of the solution , m= 100.0 mL × 1.00 g/mL = 100 g

First we have to calculate the heat gained by the solution .

$q=mc\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat of solution = $4.184 J/^oC$

$T_{final}$ = final temperature = $24.21^oC$

$T_{initial}$ = initial temperature = $23.40^oC$

Now put all the given values in the above formula, we get:

$q=100g\times 4.184 J/^oC\times (24.21-23.40)^oC$

$q=338.904 J$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 338.904 J

$moles=Molarity\times Volume (L)$

Molarity of silver nitrate = 0.100 M

Volume of silver nitrate = 50.0 mL = 0.050 L

Moles of silver nitrate = n

$n=0.100 M\times 0.050 L=0.005 mol$

$\Delta H=-\frac{338.904 J}{n}=-\frac{338.904 J}{0.005 mole}=-67,780.8 J/mol=-67.78 kJ/mol$

(1 kJ= 1000 J)

Therefore, the enthalpy change during the reaction is -67.78 kJ/mol.