Question

Two solutions, initially at 24.69°C, are mixed in a coffee cup calorimeter (Ccal = 105.5 J/°C). When a 200.0 mL volume of 0.100 M AgNO3 solution is mixed with a 100.0 mL sample of 0.100 M NaCl solution, the temperature in the calorimeter rises to 25.16°C. Determine the DH°rxn, in units of kJ/mol AgCl. Assume that the density and heat capacity of the solutions is the same as that of water.
a. -78 kJ/mol AgCl
b. -25 kJ/mol AgCl
c. -64 kJ/mol AgCl
d. -32 kJ/mol AgCl
e. -59 kJ/mol AgCl

Answer 1

Answer:The correct answer is option c.

Explanation:

$AgNO_3+NaCl\rightarrow AgCl+NaNO_3$

Specific heat of water =  4.18 J/g°C

Density of water = 1 g/ml

Mass of $AgNO_3$ solution ,m'=$1 g/ml\times 200 ml= 200 g$

Mass of $NaCl$ solution ,m''=$1 g/ml\times 100 ml= 100 g$

Since NaCl solution is present in lessor mass so, it will act limiting reagent.

Heat absorbed by the solution

$Q=mc\Delta T=(m'+m'')c\Delta T=300 g\times 4.18 J/g^oC\times (0.47^oC)=589.38 J$

Heat absorbed by the calorimeter ,$Q'=mc'\Delta T=c'\Delta T=105.5J/^oC\times (0.47^oC)=49.585 J$

Heat released during reaction,$\Delta H_{rxn}$=49.585 J+589.38 J=638.965 J

As we know that NaCl is present as limiting reagent, the product will form according to the availability of NaCl.

Moles of NaCl in 100.0 ml of 0.100 solution = 0.01 moles

1 mol of NaCl gives 1 mol of AgCl, then 0.01 mol of NaCl will give 0.01 mol of AgCl

Heat of the reaction per 1 mol of AgCl = $\frac{638.965 J}{0.01 mol}=6 63896.5 J/mol= 63.896 kJ/mol\approx 64 kJ/mol$

The heat released during the reaction that exothermic reaction's $\Delta H_{rxn}$ is negative.

Hence the correct answer is option c.