To determine how to encourage more product [tex]\( H_2CO_3 \)[/tex] to form from the reaction
[tex]\[
H_2O (g) + CO_2 (g) \rightleftharpoons H_2CO_3 (aq) + 316 \, \text{kJ}
\][/tex]
we need to consider Le Chatelier's Principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
1. Warming the reaction:
The reaction given is exothermic, as it releases 316 kJ of energy. According to Le Chatelier's Principle, if we increase the temperature (i.e., warm the reaction), the equilibrium will shift to the left to absorb the added heat. This will decrease the concentration of [tex]\( H_2CO_3 \)[/tex], hence warming the reaction will not encourage more [tex]\( H_2CO_3 \)[/tex] to form.
2. Removing [tex]\( H_2O \)[/tex]:
Removing [tex]\( H_2O \)[/tex] will shift the equilibrium towards the left to replace the removed [tex]\( H_2O \)[/tex]. This means that less [tex]\( H_2CO_3 \)[/tex] will be produced, so removing [tex]\( H_2O \)[/tex] is not the method to use.
3. Removing [tex]\( H_2CO_3 \)[/tex]:
Removing [tex]\( H_2CO_3 \)[/tex] from the reaction will shift the equilibrium to the right to produce more [tex]\( H_2CO_3 \)[/tex] to counteract the decrease in its concentration. This will encourage the formation of more [tex]\( H_2CO_3 \)[/tex].
4. Removing [tex]\( CO_2 \)[/tex] gas:
Similar to removing [tex]\( H_2O \)[/tex], removing [tex]\( CO_2 \)[/tex] gas will also shift the equilibrium towards the left to replace the removed [tex]\( CO_2 \)[/tex]. This will again decrease the production of [tex]\( H_2CO_3 \)[/tex], and thus it is not the correct method.
Therefore, to encourage more [tex]\( H_2CO_3 \)[/tex] to form, the best method is to remove [tex]\( H_2CO_3 \)[/tex].
The correct answer is:
[tex]\[ \text{remove } H_2CO_3 \][/tex]
