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At a particular temperature and pressure...

At a particular temperature and pressure, `N_(2)(g) and O_(2)(g)` react to form 4 mol of `N_(2)O`. The heat oabsorbed in this reaction is 328 kJ. What will be change in enthalpy be due to the formation of 2 mol of `N_(2)(g)` and 1 mol of `O_(2)(g)` from `N_(2)O(g)` at the same temperature and pressure?

Text Solution

Verified by Experts

The termochemical equation for the formation of `N_(2)O(g):4N_(2)(g)+2O_(2)(g) to 4N_(2)O(g),DeltaH=+328kJ`
Writing this equation in the reverse manner, we obtain,
`4N_(2)O(g)to4N_(2)(g)+2O_(2)(g),DeltaH=-328kJ`
Therefore, the thermochemical equation for the formation of 2 mol of `N_(2)(g)` and 1 mol of `O_(2)(g)` from `N_(2)O(g)`
will be: `2N_(2)O(g) to 2N_(2)(g)+O_(2)(g),DeltaH=-164kJ`
Thus, the change in enthalpy for the formation of 2 mol of `N_(2)(g)` and 1 mol of `O_(2)(g)` is -164 kJ.
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