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.

Answer 2

The ΔHrxn is  -64 KJ/mol.

The equation of the reaction is;

AgNO3(aq) + NaCl(aq) -------> AgCl(s) + NaNO3(aq)

The following information are obtainable from the reaction equation;

Total volume of solution = volume of AgNO3 solution+ volume of NaCl solution = 200.0 mL + 100.0 mL = 300.0 mL

Number of moles of AgNO3 solution = 200/1000 L ×  0.100 M = 0.02 moles

Number of moles of NaCl solution = 100/1000 L ×  0.100 M = 0.01 moles

Temperature rise = 25.16°C - 24.69°C = 0.47°C

Mass of solution = 300 g

Heat capacity of water = 105.5 J/°C

Since NaCl is the limiting reactant, 0.01 moles of AgCl is formed.

Total heat released in the reaction = Heat absorbed by calorimeter + heat absorbed by solution

ΔH = -(mcθ + cθ)

m = mass of solution

c = heat capacity of water

θ = temperature rise

Heat absorbed by calorimeter = cθ = 105.5 J/°C ×  0.47°C = 49.585 J

Heat absorbed by solution = mcθ = 300g × 4.18 J/g/°C  ×  0.47°C = 589.38 J

Total heat released in the reaction = -(49.585 J + 589.38 J)

ΔH = -638.965 × 10^-3 KJ /0.01 moles of AgCl

ΔHrxn = -64 KJ/mol

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