Question

Two solutions, initially at 24.60°c, are mixed in a coffee cup calorimeter (ccal = 15.5 j/°c). 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°rxn 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) dh°rxn = ?

Answer 1

Hey there!:

moles of AgNO3 = 100 x 0.1 / 1000 = 0.01

moles of NaCl = 100 x 0.200 / 1000 = 0.02

volume of solution = 100 + 100 = 200 mL

mass of solution = 200 x 1 = 200 g

temperature rise = 25.30 - 24.60 = 0.70ºC

Therefore:

Q = m* Cp* ΔT + Cp* ΔT =

200* 4.184 * 0.70 + 15.5 * 0.70  =

Q = 596.61 J

Given the reaction:

Q = 596.61 J

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

1 mole NaCl ------ 1 mole AgNO3 ------ 1 mole AgCl

0.02 moles NaCl ----- 0.01 moles AgNO3

So:

ΔH =  - Q / T

ΔH =  - 596.61 * 10⁻³ / 0.01

ΔH  = - 59.66 kJ

Hope that helps!