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
What do you not understand about my answer?
To determine which of the following options would result in an increase in the quantity of H2(g), we need to consider Le Chatelier's principle. According to this principle, if a system at equilibrium is subjected to a stress, the system will shift in a way that minimizes the effect of that stress.
In the given equilibrium system:
H2(g) + I2(g) <——-> 2Hl(g) + 65 kJ
1) Removing some iodine gas and removing some HI(g):
- Removing iodine gas (I2(g)) would shift the equilibrium towards the reactants side to replace the lost iodine, resulting in an increase in the quantity of H2(g).
- Removing HI(g) would also shift the equilibrium towards the reactants side to replace the lost HI, resulting in an increase in the quantity of H2(g).
Therefore, this option would result in an increase in the quantity of H2(g).
2) Decreasing temperature:
- According to Le Chatelier's principle, decreasing the temperature of an exothermic reaction shifts the equilibrium towards the products side to counteract the reduction in temperature.
- In this case, decreasing the temperature would shift the equilibrium towards the products side, resulting in a decrease in the quantity of H2(g).
Therefore, this option would not result in an increase in the quantity of H2(g).
3) Removing some iodine gas:
- As discussed earlier, removing iodine gas would shift the equilibrium towards the reactants side, resulting in an increase in the quantity of H2(g).
Therefore, this option would result in an increase in the quantity of H2(g).
4) Removing some HI(g):
- Removing HI(g) would shift the equilibrium towards the reactants side to replace the lost HI, resulting in an increase in the quantity of H2(g).
Therefore, this option would result in an increase in the quantity of H2(g).
5) Removing some HI(g) & decreasing temperature:
- As discussed earlier, removing HI(g) would shift the equilibrium towards the reactants side, resulting in an increase in the quantity of H2(g).
- However, decreasing the temperature would shift the equilibrium towards the products side, resulting in a decrease in the quantity of H2(g).
Therefore, this option would not result in an increase in the quantity of H2(g).
Based on the above analysis, options 1 (both removing some iodine gas and removing some HI(g)), 3 (removing some iodine gas), and 4 (removing some HI(g)) would result in an increase in the quantity of H2(g).
To determine which option would result in an increase in the quantity of H2(g), we need to consider Le Chatelier's principle, which states that a system at equilibrium will adjust itself to counteract any changes imposed on it.
In the given equilibrium system: H2(g) + I2(g) <——-> 2Hl(g) + 65 kJ
1) Removing some iodine gas and removing some HI(g):
According to Le Chatelier's principle, if we remove some I2(g) or HI(g), the system will try to counteract the decrease in reactants by shifting the equilibrium towards the reactant side. Since H2(g) is also a reactant, it would result in an increase in the quantity of H2(g). So, this option could potentially result in an increase in the quantity of H2(g).
2) Decreasing temperature:
Lowering the temperature of an exothermic reaction (as indicated by the negative heat of the reaction, -65 kJ) would favor the forward reaction (towards the products) to counteract the decrease in temperature. In this case, the forward reaction would lead to a decrease in the quantity of H2(g) and an increase in the quantity of HI(g) and Hl(g). So, this option would not result in an increase in the quantity of H2(g).
3) Removing some iodine gas:
As mentioned earlier, removing some I2(g) would cause the equilibrium to shift towards the reactant side, increasing the quantity of H2(g). So, this option could potentially result in an increase in the quantity of H2(g).
4) Removing some HI(g):
By removing HI(g), the system will try to counteract the decrease in reactant by shifting the equilibrium towards the reactant side. This would result in an increase in the quantity of H2(g). So, this option could potentially result in an increase in the quantity of H2(g).
5) Removing some HI(g) and decreasing temperature:
Combining the effects mentioned in options 2 and 4, removing HI(g) would shift the equilibrium towards the reactants, increasing the quantity of H2(g). However, decreasing the temperature would favor the reverse reaction (towards the reactants), decreasing the quantity of H2(g). So, this option would not result in an increase in the quantity of H2(g).
Based on the analysis above, the options that could result in an increase in the quantity of H2(g) are:
- Option 1 (both removing some iodine gas and removing some HI(g))
- Option 3 (removing some iodine gas)
- Option 4 (removing some HI(g))