Question

What is the ∆H value for this thermo chemical equation? Full question in the description box below.

Answer 1

`sf(-741color(white)(x)kJ"/""mol")`

Explanation:

`sf(M_r[C_3H_8]=44.1)`

We are told that:

1 tonne gives `sf(1.68xx10^4color(white)(x)MJ)`

`:.` `sf(cancel(10^6)color(white)(x)grarr1.68xx10^4xxcancel(10^(6))J)`

This means that:

`sf(1color(white)(x)grarr1.68xx10^4color(white)(x)J)`

We need to find the energy released by 1 mole which is 44.1 g:

`:.` `sf(44.1color(white)(x)grarr1.68xx10^4xx44.1color(white)(x)J)`

`sf(=74.088xx10^4color(white)(x)J)`

`sf(=741color(white)(x)"kJ")`

Since heat is released we can say that:

`sf(DeltaH=-741color(white)(x)"kJ""/""mol")`

Answer 2

`DeltaH=-739` `kJmol^-1`

Explanation:

First we need to calculate the number of moles of `CO_2` in `1` tonne, which is `10^6` `g`. The molar mass of `CO_2` is `44` `gmol^-1`.

`n=m/M=10^6/44=22,727` `mol`

Now the total energy produced by `1` tonne is `1.68xx10^4` `MJ` = `1.68xx10^7` `kJ`. This is energy emitted to the environment, and therefore lost from the chemical system. That means it will take a negative value when we calculate the `DeltaH`.

We need to divide this by the number of moles to find the `DeltaH` value in `kJmol^-1`.

`DeltaH=-(1.68xx10^7)/(22,727)=-739` `kJmol^-1`