Hello Mr Weng, Where exactly did you get the values for the enthalpy of formation for calcium carbonate and calcium chloride and all that in problem number 1. I couldn't find any in the data booklet :(
problem 2, the answer should be - 511.5 KJ , exothermic reaction? or does it not matter... (And also for other people getting close to the same number he gets it is probably because you have different values for the formation of the compounds (as well as him using rounded values for e.g. number of moles), I got - 553 KJ using the most recent data in the data booklet at the time, 3rd edition first assessment 2016.
Mr.Weng I still don't understand how you calculated the formulas by just looking at the questions. Could you explain it in another way or written form? Thank you.
Hi Mo Football, I'm not sure which question in particular you are referring too. Could you be more specific? With the question at 3:12 I've gone for using enthalpy of formation but I could have used bond enthalpy to work it out. If you know what the method is that they want you to use from the question eg. Bond enthalpy, Enthalpy of formation, Enthalpy of combustion ect... then you can work out the formula using the triangle I draw which has the reactants going to products. The third part of the triangle written below has the type of method or the definition really. You need to have these definitions memorized. You then use the definition to determine if the arrows go from the reactants/products to the definition, or the reverse. Eg. Enthalpy of combustion has the reactants/products going to the definition because they are burnt to CO2 and H2O. You then redraw a second set of arrows so that you go from reactant to definition to product to determine which one needs reversing. This gives you the formula as either R less P, or P less R. Hope that helps! Andrew.
Hi GummyBear121499. dH=Q/n so Q=dH x n. The enthalpy of the reaction was found (dH), but the reaction wasn't using a mole but only 0.25 of a mole so the heat (enthalpy) produced was calculated by Q = dH x 0.25. Hope that helps.
There is a small mistake 4:22 - the final result should be 65 kJ/mol.
yep
Hello Mr Weng,
Where exactly did you get the values for the enthalpy of formation for calcium carbonate and calcium chloride and all that in problem number 1. I couldn't find any in the data booklet :(
Poojitha Pai yeah i dont understand that either
Ill go ahead and second that. That information is not given in the data booklet.
I'll go ahead and third that
It is in the databooklet section 12, Selected compounds - thermodynamic data
problem 2, the answer should be - 511.5 KJ , exothermic reaction? or does it not matter... (And also for other people getting close to the same number he gets it is probably because you have different values for the formation of the compounds (as well as him using rounded values for e.g. number of moles), I got - 553 KJ using the most recent data in the data booklet at the time, 3rd edition first assessment 2016.
1:10 absolutely stellar cropping
So you use enthalpy of combustion, when measuring enthalpy of formation?
Yes, just flip it around
Mr.Weng I still don't understand how you calculated the formulas by just looking at the questions. Could you explain it in another way or written form? Thank you.
Hi Mo Football, I'm not sure which question in particular you are referring too. Could you be more specific? With the question at 3:12 I've gone for using enthalpy of formation but I could have used bond enthalpy to work it out.
If you know what the method is that they want you to use from the question eg. Bond enthalpy, Enthalpy of formation, Enthalpy of combustion ect... then you can work out the formula using the triangle I draw which has the reactants going to products. The third part of the triangle written below has the type of method or the definition really. You need to have these definitions memorized. You then use the definition to determine if the arrows go from the reactants/products to the definition, or the reverse. Eg. Enthalpy of combustion has the reactants/products going to the definition because they are burnt to CO2 and H2O. You then redraw a second set of arrows so that you go from reactant to definition to product to determine which one needs reversing. This gives you the formula as either R less P, or P less R. Hope that helps!
Andrew.
@@AndrewWeng can u explain the plus and minus sign on the triangle
@@heykiem1908 The minus sign is used because you are going in the opposite direction. Your aim is to go from R to P.
Can you explain problem two on 5:12
I don't get why for 5:33 you multiple it when you're suppose to divide.
Hi GummyBear121499. dH=Q/n so Q=dH x n. The enthalpy of the reaction was found (dH), but the reaction wasn't using a mole but only 0.25 of a mole so the heat (enthalpy) produced was calculated by Q = dH x 0.25. Hope that helps.
Great vid! Keep up the good work!
+Ackolit YT Thanks.
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