A student followed the procedure in this experiment in Experiment 28 (Heat of Ne
ID: 518954 • Letter: A
Question
A student followed the procedure in this experiment in Experiment 28 (Heat of Neutralization), using 50.0 mL of 0.500 M HC1 at 25.0 degree C and 50.0 mL of 0.500 M NaOH at 25.0 degree C in a coffee cup calorimeter, the temperature of the mixture rises to 28.2 degree C. What is the heat of reaction per mole of acid? Assume the mixture has a specific heat capacity of 4.18 J/(g middot K), heat capacity of the coffee cup is 10.7 J/K and that the density of the reactant solution is 1.00 g/mL at the end of the reaction.Explanation / Answer
Given
Volume of HCl = 50 ml = 0.05 L
Molarity of HCl = 0.5 M (mol/L
No. of moles of HCl = Molarity * Volume = 0.05 L * 0.5 mol/L = 0.025 moles
Volume of NaOH = 50 ml = 0.05 L
Molarity of NaOH = 0.5 M (mol/L
No. of moles of NaOH = Molarity * Volume = 0.05 L * 0.5 mol/L = 0.025 moles
total volume of solution = 50 ml + 50 ml = 100 ml
Given density = 1 g/ml
mass of solution = volume * density = 100 ml * 1 g/ml = 100 g
Cp of solution = 4.184 J/g.K
inital temperature = 25 C = 298 K
final temperature = 28.2 C = 301.2 K
Heat realeased = m* Cp * (T2 - T1) = 100 g * 4.184 J/g.K * ( 301.2 - 298 ) K = 1338.88 J
this is amount of heat realeased when 0.025 moles of acid reacts
Heat per mole = 1338.88 J / 0.025 mol = 53555.2 J/mol = 53.5552 KJ/mol Answer
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