Question

A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The temperature of both solutions before mixing was 23.00°C, and it rises to 26.00°C after the acid-base reaction. What is the enthalpy change for the reaction per mole of salt formed? Assume the densities of the solutions are all 1.08 g/mL and the specific heat capacities of the solutions are 4.18 J/gK. Use the correct sign.

Answer 1

Explanation:

The reaction equation will be as follows.

     $Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)$

So, according to this equation, 1 mole $Ca(OH)_{2}$ = 2 mol HBr = 1 mol $CaBr_{2}$

Therefore, calculate the number of moles of calcium hydroxide as follows.

     No. of moles of $Ca(OH)_{2}$ = $V \times Molarity$

                                    = $50 \times 0.6$

                                    = 30 mmol

Similarly, calculate the number of moles of HBr as follows.

        No. of moles of HBr = $M \times V$

                                          = $50 \times 0.6$

                                          = 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of $CaBr_{2}$ = $30 \times \frac{1}{2}$

                                                     = 15 mmol

Hence, calculate the amount of heat released as follows.

                Heat released in the reaction(q) = $m \times s \times \Delta T$

as,    m = mass of solution

and,             Density = $\frac{mass}{volume}$

or,                  mass = Density × Volume

                               = 1.08 g/ml \times (50 + 50) ml

                               = 108 g

where,    s = specific heat of solution = 4.18 j/g.k

and,        change in temperature $\Delta T$ = $(26 - 23)^{o}C$

                                                                 = $3 ^{o}C$

Hence, the heat released will be as follows.

                   q = $m \times s \times \Delta T$

                        q = $108 \times 4.18 \times 3^{o}C$

                           = 1354.32 joule

or,                        = 1.354 kJ       (as 1 kJ = 1000 J)    

Also,          $\Delta H_{rxn}$ = $\frac{-q}{n}$

                              = $\frac{-1.354}{15 \times 10^{-3}}$

                              = -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.