A common laboratory reaction is the neutralization of an acid with a base. When 31.8 mL of 0.500 M HCl at 25.0°C is added to 68.9 mL of 0.500 M NaOH at 25.0°C in a coffee cup calorimeter (with a negligible heat capacity), the temperature of the mixture rises to 28.2°C. What is the heat of reaction per mole of NaCl (in kJ/mol)? Assume the mixture has a specific heat capacity of 4.18 J/(g·K) and that the densities of the reactant solutions are both 1.07 g/mL. Enter your answer to three significant figures in units of kJ/mol.

Question

contestada

Respuesta :

SerenaBochenek SerenaBochenek · Answer 1 · 2020-06-23T08:22:53+02:00

Answer:

The correct answer to the following question will be "90.6 kJ/mol".

Explanation:

The total reactant solution will be:

[tex](31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g[/tex]

The produced energy will be:

[tex]=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K[/tex]

[tex]=450.35\times 3.2[/tex]

[tex]=1441.12 \ J[/tex]

The reaction will be:

⇒ [tex]HCl+NaOH \rightarrow NaCl+H_{2}O[/tex]

Going to look at just the amounts of reactions with the same concentrations, we notice that they're really comparable.

Therefore, the moles generated by NaCl will indeed be:

= [tex](\frac{31.8}{1000} \ L)\times (0.500 \ M \ HCl/L)\times \frac{1 \ mol \ NaCl}{1 \ mol \ HCl}[/tex]

= [tex]0.0318\times 0.500[/tex]

= [tex]0.0159 \ mole \ of \ NaCl[/tex]

Now,

= [tex]\frac{1441.12 \ J}{0.0159 \ moles \ NaCl}[/tex]

= [tex]906364.7[/tex]

= [tex]90.6 \ KJ/mol \ NaCl[/tex]