Question

Two solutions initially at 24.60°C are mixed in a coffee cup calorimeter. when a 100.0 ml volume of 0.100 m AgNO3 solution is mixed with a 100.0 ml sample of 0.200 m nacl solution, the temperature in the calorimeter rises to 25.30°C. Determine the dh for the reaction as written below. Assume that the density and heat capacity of the solutions is the same as that of water.

NaCl(aq)+AgNO3(aq) → AgCl(s)+NaNO3(aq)

a. -69 kJ
b. -35 kJ
c. -16 kJ
d. -250 kJ

Answer 1

= - 59.7 kJ/mol

Step-by-step explanation:

From the given information:

no of moles of AgNO3 = 100/1000 × 0.100 = 0.01 mol

no of moles of NaCl = 100/1000 × 0.200 = 0.02 mol

The total volume of the solution is:

= (100 + 100 ) mL

= 200 mL

mass of the solution = 200 × 1.00 = 200 g

Heat gained by the solution:

= mass × specific heat × temperature change in solution

= 200× 4.184 × (25.3 - 24.6)

= 585.76 J

Heat gained by calorimeter

= heat capacity × change in temperature of caiorimeter

= 15.5 × (25.3 -24.6)

= 10.85 J

Thus, total heat released = total heat gained

Q = 585.76 + 10.85 J

Q = 596.61 J

From above, it is obvious that AgNO3 is the limiting reagent

∴

ΔH_{rxn} = total heat released/ moles of AgNO3

= -596.61/0.01

= 59661 J/mol

= - 59.7 kJ/mol