Hey, could someone help me calculate the neutralization enthalpy using calrimetry?
Hey, the following task is to calculate the molar neutralization enthalpy between an acid and a base.
geg: V(nitric acid)=50mL
c(nitric acid)=1mol/L
V(sodium hydroxide solution)=70mL
c(sodium hydroxide solution)=1mol/L
Temperature change: 6.5K
cp=4.18KJ/g•K
m(water)=120g
Wanted: molar neutralization enthalpy.
Solution:
The formula is -(cp•m•temperature change/n)
Now all values are given except for the amount of substance (n).
How do I calculate the amount of substance?
The formula is n=c•V.
But now I use the concentration (c) and the volume (V) of the nitric acid or the sodium hydroxide solution, since they are different values for the volume?
Thank you.
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H₃O⁺ + OH¯ ⟶ 2 H₂O
So you have 50 ml of nitric acid c=1 mol/l, which is n=cV=50 mmol HNO₃. You also have 70 ml of NaOH c=1 mol/l, which is n=70 mmol. For such diluted solutions, approximately (error ⪅ 1%), 1 g ≈̂ 1 ml should hold, so we end up with m=120 g of soup, which warms by ΔT=6.5 K during the reaction.
The amount of heat is therefore ΔQ=mcΔT=3.26 kJ.
n=50 mmol of acid reacted with 50 mmol of base, the remaining 20 mmol of base had nothing to do and just watched the spectacle bored.
The molar heat of neutralization is therefore ΔQ/n = 65.2 kJ/mol. This is reasonably close to the literature value of 57 kJ/mol—more precise values are not to be expected with such a crude measurement method. However, I am surprised that the value turns out to be too high; one would actually expect a result that is too low due to the unavoidable heat losses.
Ah, so you still only have 50 mL of acid and 50 mL of base, and the remaining 20 mL of base is simply superfluous and unnecessary for the calculation. So you calculate n=c•V= 0.1 mol/L•0.50 L(acid)= amount of substance?
Ja, die Unterschußkomponente (in diesem Fall die Säure) gibt vor, wieviel Umsatz eintreten kann.
Nur die Stoffmengen zählen, die Konzentrationen braucht Du nicht zu berücksichtigen. Wenn man z.B. die HNO₃ nur 0.5 mol/l macht, aber sonst alle Zahlen gleichläßt, dann reagiert nur die halbe Stoffmenge.
Okay, wie sieht es mit unterschiedlichen Konzentrationen aus?