Question

8aq%29" id="TexFormula1" title="AgNO _3(aq) + NaCl(aq) \rightarrow AgCl(s)\ +\ NaNO _3(aq)" alt="AgNO _3(aq) + NaCl(aq) \rightarrow AgCl(s)\ +\ NaNO _3(aq)" align="absmiddle" class="latex-formula">In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃ (aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a coffee-cup calorimeter. Which of the following is the enthalpy change of the precipitation reaction represented above if the final temperature of the mixture is 21.0°C?(Assume that the total mass of the mixture is 100. g and that the specific heat capacity of the mixture is 4.2 J/(g.°C).) (A) -84 kJ/mol, (B) -0.42 kJ/mol (C) 0.42 kJ/molu (D) 84 kJ/molpx

Answer 1

Answer:

The enthalpy change during the reaction is -84 kJ/mole.

Explanation:

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

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

where,

Q = heat gained = ?

m = mass of the solution = 100.0 g

c = specific heat = $4.2 J/^oC$

$T_{final}$ = final temperature = $21.0 ^oC$

$T_{initial}$ = initial temperature = $20.0 ^oC$

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

$Q=100.0 g\times 4.2 J/^oC\times (21.0-20.0)^oC$

$Q=420 J$

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

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

where,

$\Delta H$ = enthalpy change = ?

Q = heat gained = 420 J

n = number of moles silver nitrate= n

$AgCl(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

Molarity of the silver nitrate solution = M = 0.100 M

Volume of the silver nitrate solution = V = 50.0 mL = 0.050 L

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

$\Delta H=-\frac{420 J}{0.005mole}=-84000J/mol=-84 kJ/mol$

(1 kJ = 1000 J)

Therefore, the enthalpy change during the reaction is -84 kJ/mole.