(1) What is the difference between the change in enthalpy with the circle and dash through it, and the one without it? (2) If the forwards direction of a reaction is endothermic, is it correct to say that the the reverse direction is exothermic? Basically, the reverse direction is always opposite to the forwards direction? I heard you imply that but idk if it's a rule or not.
That shift that you mentioned is precisely balanced out by another change. If you wrote out the Kp Expression if there is an equilibrium shift, you'd see there are more terms on the top or bottom. This means increasing total pressure will have the greater impact on whichever part of the expression has more partial pressure terms. These effects cancel out
6:34 sir why don't we calculate the change of moles to get then the moles that there at equilibrium? Thanks! And also, why is the enthalpy not affected by the catalyst?
On this occasion (not typical) they are asking for the % of the acid that has been used up. The change in moles (from start to equilibrium) divided by the starting moles. This gives us the percentage its changed by
thank you soo much and brilliant explanations
You're very welcome! Glad it's useful 😊
cheers, exactly what i have been craving
Enjoy the snack! 😀
@@chemistrytutor 🤣🤣🤣🤣🤣
6:12 if the ratio would've been different let's say 2H20 and 1Ch3COOH, would we divided by 2 to get the moles of the CH3COOH? Thanks!
That's right. Although to clarify, you'd divide the *change in moles* by 2 before applying that change to the acid 😀
(1) What is the difference between the change in enthalpy with the circle and dash through it, and the one without it?
(2) If the forwards direction of a reaction is endothermic, is it correct to say that the the reverse direction is exothermic? Basically, the reverse direction is always opposite to the forwards direction? I heard you imply that but idk if it's a rule or not.
1) the circle is standard conditions so standard enthalpy change
2) correct. That is indeed the rule
Doesnt increase in pressure also impact the kc constant because equilibrium shifts to the side with fewest moles of gas molecules
That shift that you mentioned is precisely balanced out by another change. If you wrote out the Kp Expression if there is an equilibrium shift, you'd see there are more terms on the top or bottom. This means increasing total pressure will have the greater impact on whichever part of the expression has more partial pressure terms. These effects cancel out
6:34 sir why don't we calculate the change of moles to get then the moles that there at equilibrium? Thanks!
And also, why is the enthalpy not affected by the catalyst?
On this occasion (not typical) they are asking for the % of the acid that has been used up. The change in moles (from start to equilibrium) divided by the starting moles. This gives us the percentage its changed by
@@soyadrink9831 😀