Question

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.

Answer 1

Answer:

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

Explanation:

The total reactant solution will be:

$(31.8 \ mL+68.9 \ mL)\times 1.07\ g/mL = 107.74 \ g$

The produced energy will be:

$=4.18 \ J/(gK)\times 107.74 \ g\times (28.2-25.0)K$

$=450.35\times 3.2$

$=1441.12 \ J$

The reaction will be:

⇒  $HCl+NaOH \rightarrow NaCl+H_{2}O$

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:

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

=  $0.0318\times 0.500$

=  $0.0159 \ mole \ of \ NaCl$

Now,

=  $\frac{1441.12 \ J}{0.0159 \ moles \ NaCl}$

=  $906364.7$

=  $90.6 \ KJ/mol \ NaCl$