Question

A strip of magnesium metal having a mass of 1.30 g dissolves in 150. ml of 5.00 M HCl (specific gravity = 1.10); products of the reaction are magnesium chloride and hydrogen gas. The HCl is initially at 22.0°C, and the resulting solution reaches a final temperature of 52.0°C. The heat capacity of the calorimeter in which the reaction occurs is 345 J/°C. Calculate ΔH (in kJ/mol) for the reaction under the conditions of the experiment, assuming the specific heat of the final solution is the same as that for water (4.184 J/g°C).

Answer 1

Step-by-step explanation:

The given data is as follows.

Mass of Mg = 1.30 g, Molarity = 5 for HCl, Volume = 150 mL

Density = 1.1 g/mL

Change in temperature (dT) =
`(52 - 22)^(o)C = 30^(o)C`

`C_(p)` = 345
`J/^(o)C`

Hence, the energy balance will be as follows.

`-Q_(rxn) = Q_(calorimeter) + Q_(water)`

`Q_(rxn) = H_(rxn)/mol`

`H_(Rxn) = Q_(Rxn)/mol`

Moles of Mg =
`\frac{mass}{\text{molecular mass}}`

=
`{1.3}{24.3}`

= 0.05349 mol of Mg

`Q_(calorimter) = C * dT`

=
`345 J/^(o)C * 30^(o)C`

= 10350 J

`Q_(water) = m * C * (T_(f) - T_(i))`

=
`150 * 1.1 * 4.184 * 30^(o)C`

= 20710.8

`-Q_(rxn)` = (20710.8 + 10350) J

`Q_(rxn)` = -31060.8 J

Therefore, calculate the value of
`\Delta H` of the reaction as follows.

`Q_(rxn) = H_(rxn)/mol`

`H_rxn = (-31060.8)/(0.05349)`

`H_(Rxn)` = -580684.24 J/mol

or,
`H_(Rxn) = -580.68 kJ/mol` (as 1 kJ = 1000 J)

Thus, we can conclude that value of
`\Delta H` of the reaction under given conditions is -580.68 kJ/mol.