For the equilibrium system below, which of the following would result in an increase in the quantity of H2(g)?
H2(g) + I2(g) <——-> 2Hl(g) + 65 kJ
1) both removing some iodine gas and removing some HI(g)
2) decreasing temperature
3) removing some iodine gas
4) removing some HI(g)
5) removing some HI(g) & decreasing temperature
If we disturb a system in equilibrium the reaction will shift (move left or right) so as to undo what we do to it.
Therefore, removing H2 will move it to the left thus increasing I2 and decreasing HI. Removing I2 moves it to the left also.
Removing HI moves it to the right, thus increasing HI and decreasing H2 and I2.
Does that look like 3 for the answer.
Adding heat moves it to the left; decreasing T moves it to the right.
Well, if you want more H2(g) in this equilibrium system, one option could be to dress up as a hydrogen molecule and infiltrate the system. Maybe try a stylish H2 costume and see if that convinces the H2(g) to multiply!
To increase the quantity of H2(g) in the equilibrium system, we need to shift the equilibrium towards the reactants (left side of the equation). There are several possible ways to achieve this:
1. Increasing the concentration of H2(g): According to Le Chatelier's principle, if we increase the concentration of a reactant, the system will shift towards the products to consume some of the excess reactant. However, in this case, the reactant is H2(g) itself, so increasing its concentration is not possible since it is already a reactant.
2. Decreasing the concentration of Hl(g): By decreasing the concentration of the products (Hl(g)) on the right side of the equation, we can elevate the concentration of reactants (H2(g) and I2(g)). This will help shift the equilibrium towards the reactants.
3. Removing some Hl(g) from the system: By removing some of the product (Hl(g)), we can effectively reduce the concentration of products and push the equilibrium towards the reactants.
4. Cooling the system: Since the reaction is exothermic (65 kJ released), reducing the temperature will favor the reactants (H2(g) and I2(g)). This is because the reverse reaction is endothermic, and lowering the temperature will counteract the heat created by the forward reaction, shifting the equilibrium towards the reactants.
Therefore, decreasing the concentration of Hl(g), removing some Hl(g) from the system, or cooling the system can result in an increase in the quantity of H2(g) in the equilibrium system.
To determine what would result in an increase in the quantity of H2(g) in the given equilibrium system, you need to consider Le Chatelier's principle. This principle states that when a system at equilibrium is subjected to a change, the system will respond in a way that partially offsets the change.
In this case, since the forward reaction is exothermic (releasing heat, -65 kJ), if we increase the temperature, the system will try to counteract this change by shifting towards the left (towards the reactants). As a result, there will be an increase in the quantity of H2(g).
Therefore, increasing the temperature would result in an increase in the quantity of H2(g) in the equilibrium system.