Question

When 50.0 mL of 0.45 M HCl was added to 50.0 mL of 0.65 M NaOH, the temperature change was 5.8 *C. Calculate the heat change for the reaction (Q_soln) and the heat of neutralization (ChangeH_neut), and the state whether the reaction is exothermic or endothermic.

Answer 1

1)

q = m⋅c⋅ΔT , where

q - heat absorbed
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as final temperature minus initial temperature

Now, the volume of the solution will be equal to the sum of the volumes of the hydrochloric acid and sodium hydroxide solutions

V_sol = V_HCl + V_NaOH

V_sol = 50. mL+50. mL = 100. mL = 100 g ------------ [since water density = 1g/mL and specific heat = 4.184 J/g^oC]

Use the density of the solution to find its mass

now Q_soln = 100g x 4.184 J/g^oC x 5.8 ^oC

Q_soln = 2426.72 J

2) ΔH_neut = -Q_soln ÷ n(H₂O_(l))

neuralisation is an exothermic reaction, it releases heat, so ΔH_neut must be negative.

moles HCl = 50.0 mL x 10^-3 L x 0.45 M = 0.0225 mol

moles naOH = 50.0 mL x 10^-3 L x 0.65 M = 0.0325 mol

Since you have fewer moles of hydrochloric acid, the acid will act as a limiting reagent, i.e. it will determine how many moles of sodium hydroxide actually take part in the reaction.

More specifically, the acid will be completely consumed by the reaction, and only 0.0225 moles of sodium hydroxide will react.

The heat given off by this reaction per mole will thus be

2426.72 J / 0.0225 mol = 107854.222222 J/mol

ΔH_neut = −107.85 kJ/mol         --------------------- [Since 1kJ = 1000 J]

We have final answers:    Q_soln = 2.43 kJ & ΔH_neut = −107.85 kJ/mol

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