Question

You place 40.6 ml of 0.659 M NaOH in a coffee- cup calorimeter at 25.00°C and add 71.3 ml of 0.659 M HCl, also at 25.00°C. After stirring, the final temperature is 28.57°C. [Assume the total volume is the sum of the individual volumes and that the final solution has the same density (1.00 g/ml) and specific heat capacity (4.184 J/gK)]. Calculate the change in enthalpy (LaTeX: \DeltaΔH) of the reaction in kJ/mol of water formed. Enter to 1 decimal place.

Answer 1

Answer:

ΔH of reaction is 62,3 kJ/mol

Explanation:

For the reaction:

HCl + NaOH → H₂O + NaCl

The heat transfer in the reaction (q) =

q = Specific heat × mass × ΔT

The mass is 40,6 mL + 71,3 mL × (1g/mL) = 111,9 g

And ΔT is 28,57 - 25,00 = 3,57°C

q = 4,184 J/gK × 111,9 g × 3,57°C = 1671 J

The moles of NaOH are:

0,0406 L×0,659 M = 0,0268 moles

And the moles of HCl are:

0,0713 L ×0,659 M = 0,0470 moles

The moles of reaction are the moles of NaOH because are the limiting reactant. 0,0268 moles

ΔH of reaction is q/moles of reaction. Thus:

ΔH of reaction =  1671J/0,0268 moles = 62350 J/mol ≡ 62,3 kJ/mol

I hope it helps!

Answer 2

The change in enthalpy for the reaction of formation of water is 62.367 kJ/mol.

The reaction of HCl and NaOH is given as:

$\rm HCl+NaOH\to\;H_2O+NaCl$

The equal moles of HCl and NaOH are requires for the reaction.

Computation for the change in enthalpy of the reaction

The change in enthalpy ($\Delta H$) for the reaction is given by:

$\Delta H=\rm \dfrac{mass\;\times\;specific\;heat\;\times\;change\;in\;temnperature}{moles}$

The moles of the limiting reactants in the equation are given as:

$\rm Moles=Molarity\;\times\;Volume$

• The moles of HCl are:

$\rm Moles\;HCl=0.0406\;\times\;0.659\\Moles\;HCl=0.0470\;mol$

• The moles of NaOH in the solution are:

$\rm Moles\;NaOH=0.0713\;\times\;0.659\\Moles\;NaOH=0.0268\;mol$

The moles of NaOH are less, thus, it is the limiting reactant

The mass of the reaction mixture is given as:

$\rm Mass=Volume\;\times\;Density$

Substituting the values for the mass:

$\rm Mass=Volume\;HCl+Volume\;NaOH\;\times\;density\\Mass=71.3\;+\;40.6\;\times\;1\;g\\Mass=111.9\;g$

The value of specific heat for the reaction is 4.184 J/gK

The change in the temperature is given as:

$\rm Change\;in\;temperature=Final \;temperature- Initial\;temperature\\ Change\;in\;temperature=28.57-25^\circ C\\ Change\;in\;temperature=3.57^\circ C$

Substituting the values for the change in enthalpy:

$\rm \Delta \textit H=\dfrac{111.9\;\times\;4.184\;\times\;3.57}{0.0268} J/mol\\\Delta \textit H=62367 J/mol\\\Delta \textit H=62.367\;kJ/mol$

The change in enthalpy for the reaction of formation of water is 62.367 kJ/mol.

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