Question

How do I determine the standard enthalpy of combustion for magnesium in hydrochloric acid?

Reaction 1
mass= 0.13g
Starting temp= 23.0 C
Highest temp= 34.8 C
50mL of HCl
Reaction 2
Mass= 0.97
Starting temp= 23.0 C
Highest temp= 38.0 C
50mL of HCl

Answer 1

There isn't a way, as far as I can see, of finding the standard enthalpy of combustion from here. I have found the standard enthalpy of reaction below, although the numbers only seem to fit for the first experiment. Maybe I've misunderstood though;

regardless, the enthalpy of reaction is `-461" kJ mol"^-1`

Explanation:

Consider reaction 1

We will make a fairly reasonable assumption here, that the HCl is basically water. This will allow us to use water's density, and specific heat capacity.

Since `1ml=1cm^3`

`"volume of HCl"=50cm^3`

The density of water is `1" g cm"^-3`

`:."mass HCl"=1xx50`
`=50g`

Now we use `Q=mcDeltaT`
`Q=50xx-4.18xx(34.8-23)`
`=-2466.2J` (write all your sig figs at this point).
`=-2.4662kJ`

To find the enthalpy of reaction, we need to divide by the moles of magnesium (the unit is kJ per mole).

`"mol Mg"=0.13/24.3`
(I'm taking the `A_r` of magnesium to be 24.3)

`"mol Mg"=5.350xx10^-3" mol"`
I've only written 4sf, so keep this number in your calculator; use the ANS button.

So the enthalpy of reaction:

`:. Delta_rH=-2.4662/(5.350xx10^-3)`
`=-461" kJ mol"^-1 " " (3"sf")`

Here we round our answer.

Consider reaction 2
You should do the same thing again for this question. I have included my calculation, but the answer is off by a long shot. I don't know if this is my calculation or an error with the question - either way, the first reaction enthalpy we calculated seems more accurate.

`Q=mcDeltaT`
`=50xx-4.18xx(38-23)`
`=-3135J`
`=-3.135kJ`

`"mol Mg"=0.97/24.3`
`=0.0399" mol"`

`Delta_rH=-3.135/0.0399`
`=-78.57" kJ mol"^-1`
(see what I mean?)