Answer to Question #180787 in General Chemistry for Danika

Question

In an adiabatic calorimetry experiment, 100 mL of 0.5 M HCl was added to 100 mL 0.5 M NaOH to produce 200 mL 0.25 M NaCl. A temperature increase of 3.38 C was measured.

a) calculate the heat of reaction associated with this process at 25 C

b) use this value of ^H to calculate the capacity of the system. Note the density of a 0.25 molar NaCl solution is 1.02 gm/ml.

Answer 1

NaOH + HCl ------> NaCl + H₂O

HCl: -166.19 kJ

NaOH: -469.07 kJ

NaCl: -406.77 kJ

ΔH^º = ∑Δv_pΔH^º_f(products) - ∑Δ v_rΔH^º_f(reactants)

= (-406.77) - (-469.07 -166.19)

=+228.49kJ

q=mass*speed of light*change in temperature

mass= 250 moles * 1.02g/mol

= 255g

q=255 g * 2.99792*10^8m/s * (273+3.38)K

= 2.113 * 10^13 J/K