. H2(g) + I2(g) ↔ 2HI(g) ∆H = +52.7 kJ/mol
a) increase [H2]
b) increase [HI]
c) increase the pressure by lowering the volume
d) add a catalyst
You posted an equation with actions but no question. I assume you want to know which way the reaction shifts (Le Chatelier's Principle).
Here's the way you know. In simple but non-esoteric terms Le Chatelier's Principle says "when a system in equilibrium is exposed to a stress it will shift so as to undo what we've done to it. Keep that in mind.
a) increase [H2]
If you increase H2 it will shift so as to get rid of H2. How can it do that? It can move to the right so as to increase HI, decreasing H2, and decreasing I2 in the process.
b) increase [HI] same logic as a which means it will shift to the left.
c) increase the pressure by lowering the volume
Lowering the volume increases pressure. The reaction will shift to the side with fewer mols of gas for increasing P. Since you have 2 mols gas on the left and 2 mols on the right there will be no shift.
d) add a catalyst Adding or taking away of a catalyst doesn't change the equilibrium so there is no shift. Adding or removing a catalyst affects how FAST equilibrium is reached but it does not affect the POINT of equilibrium.
I love these Le Chatelier's problems. They really are simple to answer but they always give students so much trouble. I never understood why. The answer always is that the equilibrium just "does away what we've done to it".
To determine the effect of each change on the equilibrium system H2(g) + I2(g) ↔ 2HI(g), we can refer to Le Chatelier's principle, which states that a system at equilibrium will respond to a stress by trying to minimize the effect of the stress.
a) Increasing [H2]:
If you increase the concentration of H2, according to Le Chatelier's principle, the system will try to reduce the stress by favoring the forward reaction. This means that the equilibrium will shift to the right, producing more HI until a new equilibrium is established.
b) Increasing [HI]:
If you increase the concentration of HI, the system will try to reduce the stress by favoring the reverse reaction. The equilibrium will shift to the left, decreasing the concentration of HI until a new equilibrium is established.
c) Increasing the pressure by lowering the volume:
If you decrease the volume of the system, the pressure increases. According to Le Chatelier's principle, the system will try to reduce the stress caused by the increased pressure. In this case, the system will shift in the direction that produces fewer moles of gas. The forward reaction produces 2 moles of gas (H2 + I2), while the reverse reaction produces 1 mole of gas (2HI). Therefore, the equilibrium will shift to the right, producing more HI until the pressure decreases and a new equilibrium is reached.
d) Adding a catalyst:
Adding a catalyst does not affect the position of the equilibrium. A catalyst increases the rate of both the forward and reverse reactions equally but does not shift the equilibrium position.
In summary:
a) Increasing [H2] shifts the equilibrium to the right.
b) Increasing [HI] shifts the equilibrium to the left.
c) Increasing the pressure by lowering the volume shifts the equilibrium to the right.
d) Adding a catalyst does not affect the equilibrium position.
To determine which factors will shift the equilibrium of the given reaction, we need to consider Le Chatelier's principle. According to this principle, if a system at equilibrium is subjected to a change, it will try to oppose that change and establish a new equilibrium.
a) Increase [H2]: According to Le Chatelier's principle, if we increase the concentration of one of the reactants, in this case, H2, the system will try to shift towards the products to balance the change. Therefore, the equilibrium will shift to the right, favoring the formation of more HI.
b) Increase [HI]: If we add more HI to the system, the equilibrium will once again shift to the left (opposite direction) to consume the excess HI. This is because the reaction is exothermic, meaning it releases heat, therefore, consuming more HI will help alleviate the stress.
c) Increase the pressure by lowering the volume: Since there are fewer gaseous moles on the left side of the equation compared to the right side (1+1 = 2), increasing the pressure by decreasing the volume will favor the side with fewer gaseous moles. Therefore, the equilibrium will shift to the right to produce more HI.
d) Add a catalyst: A catalyst doesn't affect the position of the equilibrium; instead, it speeds up the rate of the reaction by providing an alternative reaction pathway with lower activation energy. Therefore, adding a catalyst will not shift the equilibrium of the reaction.
In summary, increasing [H2], increasing [HI], and increasing the pressure by lowering the volume will shift the equilibrium to favor the formation of more HI. Adding a catalyst, however, will have no effect on the equilibrium position.