Answer to Question #153870 in General Chemistry for Sabrina Ahmed

Question

Calculate the heat given off or absorbed by the neutralization reaction

info given

Ba(OH)2 volume-50.0 mL and molarity-1.000 M

HCl(aq). volume-50.00 mL and molarity-1.000 M.

Initial temperature: 20 degrees celsius Final temperature: 26.81 degrees celsius

Answer 1

Assuming the density and specific heat capacity of the resulting solution to be the same as of the pure water, the total mass of the solution is:

$m=(50.0mL+50.0mL)\times1.00g/mL=100g$

Therefore,

$$Q=cm\Delta{T}=4.18J/g\cdot^{\circ}{C}\times100g\times(26.81^\circ{C}-20^\circ{C})=2.85kJ$$

Since the temperature of the solution has increased during the reaction, the heat was given off.

The given values of molarity do not affect the actual heat that was involved during the reaction, but may be used to calculate the heat per one mole of the reactant:

Ba(OH)₂ + 2HCl --> BaCl₂ + 2H₂O

Both solutions have the same volume and molarity, hence HCl is the limiting reactant because 2 moles of HCl are required to neutralize 1 mole of Ba(OH)₂.

$n(HCl)=1.000M\times0.0500L=0.0500mol$

According to the equation, the amount of Ba(OH)₂ that was used up during the reaction is:

$n(Ba(OH)_2)=\frac{n(HCl)}{2}=0.0250mol$

$Q=\frac{2.85kJ}{0.0250mol}=114kJ/mol$

Answer: 2.85 kJ given off, or 114 kJ per mole Ba(OH)₂