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

Step-by-step 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 * Molarity`

=
`50 * 0.6`

= 30 mmol

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

No. of moles of HBr =
`M * V`

=
`50 * 0.6`

= 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of
`CaBr_(2)` =
`30 * (1)/(2)`

= 15 mmol

Hence, calculate the amount of heat released as follows.

Heat released in the reaction(q) =
`m * s * \Delta T`

as, m = mass of solution

and, Density =
`(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 * s * \Delta T`

q =
`108 * 4.18 * 3^(o)C`

= 1354.32 joule

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

Also,
`\Delta H_(rxn)` =
`(-q)/(n)`

=
`(-1.354)/(15 * 10^(-3))`

= -90.267 kJ/mol

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