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ICE Table Calculating Equilibrium Constant 

Old School Chemistry
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17 сен 2024

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Комментарии : 25   
@katiep.9736
@katiep.9736 3 года назад
You’re so helpful! I have an exam in a couple of days and you are a lifesaver! Thanks for uploading.
@OldSchoolChemistry
@OldSchoolChemistry 3 года назад
Hi Katie, I’m so glad! Good luck on your test😊
@mgloriabarela4261
@mgloriabarela4261 2 года назад
I want to thank you for making this equilibrium video as well as the concepts of how to get to equilibrium by demonstrating the “seesaw” visual. I’m the type of student (sometimes, not always) who needs that extra explanation or “visual” to get the point of steps. From the bottom of my heart, I can’t thank you and the teacher who put the extra effort for students such as my self to get the grasp of what we’re actually learning. It’s teachers like you and alike who are the real MVP’s! ❤ God bless you!
@OldSchoolChemistry
@OldSchoolChemistry Год назад
H MGloria, I am so glad the videos were helpful. It makes me very happy. I am proud of your extra work and effort searching for videos to better understand the content. God bless you!! Thanks:)
@fana1395
@fana1395 Год назад
Thank you so much! Your videos are great and so helpful
@OldSchoolChemistry
@OldSchoolChemistry Год назад
Hi Fana, I am so glad! Thank you for your comment;)
@charank.7667
@charank.7667 Год назад
Hi, I am not sure if I am understanding this incorrectly but in my calculator when I input (0.035)(0.035)/(0.015) I am getting 0.81666...7, I might have missed a step or misunderstood, could you explain if I missed a calculation or misunderstood the calculation pls. Btw, love your videos, never have I understood chem so well, thank you using your time to make these videos.
@OldSchoolChemistry
@OldSchoolChemistry Год назад
Hi Charank, You are correct! Thank you for catching that. I must have made a calculator error. Thanks again!
@charank.7667
@charank.7667 Год назад
@@OldSchoolChemistry No problem, thank you so much for clarifying!
@llennzo
@llennzo 4 месяца назад
@@OldSchoolChemistryso it’s going to be reactant favoured? Because it’s less than 1?
@OldSchoolChemistry
@OldSchoolChemistry 4 месяца назад
@@llennzo Yes:) Well done.
@alexanderacosta911
@alexanderacosta911 2 года назад
Awesome videos
@OldSchoolChemistry
@OldSchoolChemistry 2 года назад
Thanks A costa! Please subscribe if you have not already. Thanks!! :)
@jennipherchanda6266
@jennipherchanda6266 Год назад
thank you so much 😘🥰 you are a beautiful soul
@OldSchoolChemistry
@OldSchoolChemistry Год назад
Hi Jennipher, thank you for your comment:) I appreciate you!💗
@jennipherchanda6266
@jennipherchanda6266 Год назад
@@OldSchoolChemistry they brought the exact question in the exam yesterday 😍😍💃💃💃💌
@OldSchoolChemistry
@OldSchoolChemistry Год назад
@@jennipherchanda6266 Hooray! And you were completely prepared. Way to go!!!!!
@chiararomano1818
@chiararomano1818 Год назад
K is not unitless is it? Here wouldn’t it have units of M? In general, it would depend on the degree of the numerator minus the degree of the denominator.
@OldSchoolChemistry
@OldSchoolChemistry Год назад
It is a ratio of product and reactant concentrations which are both M. You are correct, based on exponents, mathematically it can have a final M to the x power. However, because it is concentration divided by concentration it is considered unitless. It is just a ratio of concentrations at equilibrium for K. Thanks!
@chiararomano1818
@chiararomano1818 Год назад
@@OldSchoolChemistry Thanks. I didn’t know that. The constants for reaction rates carry units, I just assumed the equilibrium constants would as well.
@OldSchoolChemistry
@OldSchoolChemistry Год назад
@@chiararomano1818 Hi Chiara, It is one of the main differences between the equilibrium constant (K) and rate constant (k). It is so great you noticed that!
@ianfernandez8103
@ianfernandez8103 2 года назад
I think I might have done the calculations wrong; I keep getting .082 for my equilibrium constant value
@OldSchoolChemistry
@OldSchoolChemistry 2 года назад
Hi Ian, you are correct! Thank you for catching that. I somehow accidentally reciprocated the answer. Yes it is 0.082. Thanks!
@ianfernandez8103
@ianfernandez8103 2 года назад
@@OldSchoolChemistry Thank you so much btw your awesome, without you I don't think I would be succeeding in chem 2!
@OldSchoolChemistry
@OldSchoolChemistry 2 года назад
Good luck in your class 😊
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