C2H6(g) <=> C2H2(g) + 2 H2(g)
The ΔH° for the reaction above is +312 kJ. The system is initially at equilibrium. In which direction will the reaction shift in each of the following situations?
e) The temperature is decreased.
g) He is added at constant volume.
h) He is added at constant pressure.
(Possible Answers: forward, reverse, no shift)
My Work:
I did a-d and f, these are the ones that I have no clue how to do.
I do not understand how the +312 kJ affects anything. I thought decreasing temperature wouldn't cause a shift, but the assignment says I'm wrong. Why does it shift one way or the other?
Also, how can helium just be added to the reaction? And how does that effect the rest of the compounds involved?
Since the dH reaction is +312 kJ, that means the reaction is endothermic so I would write it as
C2H6 + heat = C2H2 + 2H2
So if T is decreased, rxn shift to the left. (Rxn uses heat to go to the right; take heat away it moves to the left).
Add He at constant volume. If volume does not change, there is no change in concentration (still same n and L so mols/L = constant) and there is no shift.
If He added at constant pressure, that means volume must increase to keep P from rising. If volume increases, reaction shifts to larger number of molecules (to the right).
Ah, the wonderful world of chemistry! Let's dive in and explore these questions with a little humor, shall we?
e) The temperature is decreased: Imagine the reaction as a bunch of little particles having a party. Now, picture this party happening in a room with no air conditioning. It's getting hot! So, what do the particles do to cool off? Well, they start sweating...I mean, they start shifting in different directions! In this case, when you decrease the temperature, those sweaty particles will move towards the side of the reaction that will produce heat, just to warm things up. So, the reaction will shift in the...*drumroll please*...reverse direction!
g) He is added at constant volume: Oh, look who decided to join the party! It's our helium friend, Mr. Noble Gas! He's a pretty chill and easy-going guy, you know? When you add him to the party, he doesn't really interact much with the other guests - he just kinda floats around, minding his own business. So, his presence won't really affect the forward or reverse direction of the reaction. No shift for you, Mr. Helium!
h) He is added at constant pressure: Now, this is a bit different from the previous scenario. Imagine the party happening in a confined space, with a constant pressure (let's call it "Party Goer Pressure"). When Mr. Helium comes in, he takes up some space, like a balloon. He's like that one friend who always brings too many balloons to parties, you know? So, with this addition of extra volume, the particles get kind of crowded and don't have as much space to move freely. To cope with this, the reaction will shift in the...*one more drumroll, please*...reverse direction again!
So, to summarize: when you decrease the temperature, the reaction shifts in the reverse direction. When helium is added at constant volume or constant pressure, there is no shift in the reaction. Hope this helps lighten up your chemistry journey a bit!
To understand why the direction of the reaction will shift in each situation, we need to consider Le Chatelier's principle. According to this principle, a system at equilibrium will respond to changes in temperature, pressure, or concentration in a way that minimizes the effect of that change.
e) The temperature is decreased: When the temperature is decreased, the reaction will shift in the exothermic direction to counteract the decrease in temperature. This means the reaction will shift to the right to produce more heat, resulting in more C2H2 and H2 and less C2H6.
g) He is added at constant volume: Helium (He) in this case acts as an inert gas and does not participate in the reaction. Adding an inert gas at constant volume will not have an effect on the equilibrium position of the reaction. Therefore, there will be no shift in the reaction.
h) He is added at constant pressure: In this situation, adding helium at constant pressure will increase the total pressure of the reaction mixture by increasing the number of moles of gas. According to Le Chatelier's principle, the system will shift to reduce the pressure. Since the reaction produces more moles of gas on the right side, the reaction will shift to the right, producing more C2H2 and H2 and consuming more C2H6.
It's important to note that the +312 kJ given for the reaction is the enthalpy change (ΔH°) of the reaction, which indicates that the reaction is endothermic. This information does not influence the direction of the shift in any of the situations mentioned above. It simply tells us that the forward reaction is favored at higher temperatures.
To determine in which direction the reaction will shift, we need to consider Le Chatelier's principle, which states that when a system at equilibrium is subjected to a change in temperature, pressure, or concentration, it will shift in a way that counteracts that change.
e) The temperature is decreased:
When the temperature is decreased, the system will shift in the direction that produces heat. This is because the reaction is exothermic, meaning it releases heat. By shifting in the forward direction (from left to right), the reaction will produce more heat, which helps counteract the decrease in temperature. Therefore, the system will shift in the forward direction.
g) He is added at constant volume:
When helium (He) is added at constant volume, it does not directly affect the reaction. Helium is an inert gas and does not participate in the chemical reaction. Therefore, the addition of helium at constant volume will not cause a shift in the equilibrium position. The answer is "no shift."
h) He is added at constant pressure:
When helium is added at constant pressure, it increases the total pressure in the system. According to Le Chatelier's principle, the system will shift in a direction that reduces the pressure. In this case, adding helium at constant pressure favors the side of the reaction with fewer gas molecules (C2H6 and C2H2). As C2H6 and C2H2 have one gas molecule each, and H2 has two gas molecules, the system will shift in the reverse direction (from right to left) to decrease the total pressure. Therefore, the system will shift in the reverse direction.
The +312 kJ value mentioned as the ΔH° for the reaction does not directly affect the shifting of the reaction. It is simply the enthalpy change associated with the reaction and provides information about the energy released or absorbed during the reaction. The shifting of the reaction depends on changes in temperature, pressure, or concentration, as explained above.