Calculate the molar enthalpy change for the reaction below.

Question

Use the following information to answer the next question.

Ba(NO3)2 is a barium compound that is highly soluble in water. When it is added to a solution of sodium hydroxide, a precipitate forms:	Ba(NO3)2(aq) + 2NaOH(aq) à Ba(OH)2(s) + 2 NaNO3(aq)

A 40.0 mL sample of 0.100 mol/L Ba(NO₃)₂BaSO₄(aq) is mixed with 20.0 mL of excess NaOH(aq) in calorimeter. During the reaction, there is a rise in temp by 5.00 ^oC. Calculate the molar enthalpy change for the above reaction.

Answer 1

First, we should find the heat of the reaction. The appropriate equation is q = mC∆T

q = heat = ?

m = mass = 40 ml + 20 ml = 60 ml x 1 g/ml = 60 g (note: we are assuming a density of 1 g/ml for water and we are ignoring the masses contributed by the reagents and the products of the reaction. These are common assumptions to make in this type of problem, especially in Intro Chem.)

C = specific heat = 4.184 J/gº (for water; another assumption)

∆T = change in temperature = 5.00ºC

q = (60 g)(4.184 J/gº)(5º) = 1255 J This is the heat (enthalpy) of reaction.

Since they ask for the MOLAR enthalpy, we need to find the moles of Ba(OH)₂ formed).

moles Ba(NO₃)₂ = 0.040 L x 0.1 mol /L = 0.004 moles

Molar enthalpy of reaction = 1255 J / 0.004 moles = 313,750 J/mol = 314 kJ/mol