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Funny Expt 3 from STAWA
- $N_{2}O_{4} \rightleftharpoons 2NO_{2}$
- $\Delta H= +57kJ$
This implies that the forward reaction is endothermic.
Hence, according to Le Chatelier's principle, placing such a system in hot water means the imposed change is an increased system temperature. Hence this change will be counteracted by reducing the temperature, thus the endothermic forward reaction will be favoured as it absorbs energy, reducing the temperature. Hence we will observe the system become more brown as more NO2 is produced.
When temperature increases
- Both reaction rates increase.
- The higher $E_{a}$ of the forward (endothermic) reaction means that this reaction is affected more by changes in temperature, i.e. it is more sensitive.
- Explanation: Maxwell-Boltzmann diagram
- Proportional increase of particles now able to undergo forward reaction compared to the increase for the reverse reaction is significantly more, hence the rate of the forward reaction increases over the reverse reaction rate.
- e.g for reverse, number of particles is doubled, but for forward, number of particles is 8 times more!
- TLDR proportionally more particles will gain $E_{a}$ or better for the endothermic reaction
This results in the endothermic reaction rates increasing more, hence this reaction is the one that is favoured.
See two graphs: Rate and concentration
- Note: due to molar ratios, there are two moles of $NO_{2}$ produced per mole of $N_{2}O_{4}$ consumed.
- Hence, $[NO_{2}]$ is increased by 2x, $[N_{2}O_{4}]$ is decreased by x.
- Can't be both increased, that would be magic!
NB
Regardless of whether you increase or decrease temperature, the endothermic reaction is always going to be affected the most. It is more sensitive to changes in temperature since it has the highest $E_{a}$.
Catalyst?????????
- No effect on equilibrium. Rates are increased equally.
