if the reaction for the precipitate of lead iodide is eqilibium, and the factor we change is temperature what is the out come
When the temperature increases during a reaction at equilibrium, the equilibrium favors the endothermic side. When the temperature decreases during a reaction at equilibrium, the equilibrium favors the exothermic side. This is because when the temperature increases, the endothermic side will use up the heat, and when the temperature decreases, the exothermic side will release more heat. So it all depends on how you change the temperature. Precipitation itself is an exothermic reaction, because when two aqueous substances combine to form a solid substance and an aqueous substance, the entropy decreases (liquids particles move faster than solid particles) and heat is released. Hope this helps!
if the temperature increases during the process?
Then the reaction will shift in the endothermic direction, which would be the reverse of the precipitation reaction (since it is exothermic). I realize I made a mistake by saying endothermic "side" earlier, when a better way to say it would be "direction." Anyways, since it shifts to the reverse of the precipitation reaction, we will end up with more reactants. Does that make sense?
i want to know if the mass will be high or low of the precipitate?
is this statement correct? According to me, I think that the mass of the precipitate will be more when it is heated. It is because mostly reactions occur faster when they are heated.
No, sorry :( An endothermic reaction, which requires heat, will certainly occur faster when it is heated. An exothermic reaction, which releases heat, will not occur faster when heated. Since the temperature is increasing, the equilibrium will favor the direction in which heat is being absorbed. That is, the endothermic direction. When two aqueous substance (substances immersed in liquid) combine to form a solid substance, the entropy decreases, causing heat to be released. This is because liquid particles move faster, or have more entropy, than solid particles. Heat is released, so it is exothermic. Since a precipitation reaction occurs when two aqueous substances form a solid substance and an aqueous substance, a precipitation reaction is an exothermic reaction. Since it is releasing heat, adding more heat will not speed up this reaction. Since this is an equilibrium reaction, it also occurs in the opposite direction. When a solid substance and an aqueous substance combine to create two aqueous substances, the entropy increases and heat is absorbed. Therefore, the reverse of a precipitation reaction is endothermic, or requires heat. If the temperature is increased, this side of the reaction will be favored. That is, more precipitate and aqueous solution will react. Based on this, do you think the mass of the precipitate will increase or decrease?
reply fast please :(
mass will increase..
No, sorry. If you have some precipitate and have it react with an aqueous substance to form two new aqueous substances, how will the mass of the precipitate be affected?
actuallu this reaction is for theoritical yield. PErcentage yield actually. so... here we have to find how is percentage yiel affected by temperature for that we have to measure the mass of the actual yield. so our actual yield is mass of precipitate
Theoretical yield is the amount based on calculations. Actual yield is the actual amount found in a lab environment. Percent yield is actual yield/theoretical yield * 100. If the temperature is increased, the equilibrium will favor the endothermic direction, and the amount of precipitate will decrease. So, the actual yield will be less than the theoretical yield and the percent yield will be less than 100. Does that make sense?
yes. it does. but still mass will increase right?
If you have some precipitate, and it reacts with an aqueous substance, what will happen to the mass of the precipitate? Increase or decrease?
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