Use of the Data Booklet is relevant to this question.
A student mixed 25 cm of 0.10 mol dm' sodium hydroxide solution with 25 cm of 0.10 mol dm3
hydrochloric acid and noted a temperature rise of 2.5°C.
What is the enthalpy change of the reaction per mole of NaOH?
A -209 kJ mol-1
B-104.5 kJ mol-1
C -209 Jmol-1
D -522.5 J mol-1
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Answer:
Correct option is B)
Calculate heat absorbed by the water:
(Assume density of solutions = 1.00 g/cm^3 and specific heat of solution is the same as water)
q = m c (T2-T1)
q = 50 g (4.184 J/gC) ( 30-15 C) = 3138 J
Heat released by neutralization reaction = -3138 J
Moles HCl used = 0.0250 L X 4.00 mol/L = 0.100 mol
Delta H = -3138 J / 0.100 mol = -3.14X10^4 J = -31.4 kJ/mol
B is the correct answer.
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