Question

100 mL of a 0.300 M solution of AgNO3 reacts with 100 mL of a 0.300 M solution of HCl in a coffee-cup calorimeter and the temperature rises from 21.80 °C to 23.20 °C. Assuming the density and specific heat of the resulting solution is 1.00 g/mL and 4.18 J/g ∙ °C, respectfully, what is the ΔH°rxn?

Answer 1

Answer:

ΔH°rxn = 39013.33 J/mol = 39.013 kJ/mol.

Explanation:

• We can calculate the amount of heat (Q) released from the solution using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of AgNO₃)/(Volume of the solution (L)).

∴ no. of moles of AgNO₃ = (M)(Volume of the solution (L)) = (0.3 M)(0.1 L) = 0.03 mol.

∴ ΔH°rxn = Qrxn/(no. of moles of AgNO₃) = (1170.4 J)/(0.03 mol) = 39013.33 J/mol = 39.013 kJ/mol.