Why there is no change in solubility of nacl with increase in temprature?
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For 9-10 class
Nacl Is An Ionic Bond Which Have Strong Intermolecular Force So, Nacl has no Change in solubility
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For Higher Classes
One way to understand solubility is to start with the Clausius -Clapyron equation (CC), which despite approximations involved is a good description the vapour pressure over various solids and liquids. For the ideal solution
ln(p2p1)=−ΔHvapR(1T2−1T1)ln(p2p1)=−ΔHvapR(1T2−1T1)
where p is the vapour pressure, T the temperature and ΔHΔH the enthalpy change between states 11 and 22.
The heat of sublimation is ΔHsub=ΔHvap+ΔHfusΔHsub=ΔHvap+ΔHfus and calculating the pressure over the pure solid form of the solvent gives
ln(p2p∗1)=−ΔHsubR(1T2−1T1)ln(p2p1∗)=−ΔHsubR(1T2−1T1)
and so subtracting these two equations gives
ln(p∗1p1)=ΔHfusR(1T2−1T1)ln(p1∗p1)=ΔHfusR(1T2−1T1)
If it is assumed that Raoult’s Law applies then p1=xp∗1p1=xp1∗ where x is the mole fraction of the solute. Substituting into the last equation gives
ln(x)=−ΔHfusR(1T2−1TM)ln(x)=−ΔHfusR(1T2−1TM)
where now TMTM is the melting temperature of the pure solute. As ΔH/TMΔH/TM is a constant then the mole fraction and hence solubility in solution is
ln(x)∝−ΔHfusR(1T2)ln(x)∝−ΔHfusR(1T2)
which shows that the mole fraction at temperature T varies as
xT∝exp(−ΔHfusRT)xT∝exp(−ΔHfusRT)
and so the solubility rises with increase in temperature. Different species will rise more or less slowly depending on their heat of fusion as ΔHfus/RΔHfus/R.
Using data for NaCl shows that the mole fraction hardly varies between 200200 to 400400 K, whereas there is a huge increase for NaNO3NaNOX3 under the same conditions.
Nacl Is An Ionic Bond Which Have Strong Intermolecular Force So, Nacl has no Change in solubility
---@@@@---------@@@@@-------
For Higher Classes
One way to understand solubility is to start with the Clausius -Clapyron equation (CC), which despite approximations involved is a good description the vapour pressure over various solids and liquids. For the ideal solution
ln(p2p1)=−ΔHvapR(1T2−1T1)ln(p2p1)=−ΔHvapR(1T2−1T1)
where p is the vapour pressure, T the temperature and ΔHΔH the enthalpy change between states 11 and 22.
The heat of sublimation is ΔHsub=ΔHvap+ΔHfusΔHsub=ΔHvap+ΔHfus and calculating the pressure over the pure solid form of the solvent gives
ln(p2p∗1)=−ΔHsubR(1T2−1T1)ln(p2p1∗)=−ΔHsubR(1T2−1T1)
and so subtracting these two equations gives
ln(p∗1p1)=ΔHfusR(1T2−1T1)ln(p1∗p1)=ΔHfusR(1T2−1T1)
If it is assumed that Raoult’s Law applies then p1=xp∗1p1=xp1∗ where x is the mole fraction of the solute. Substituting into the last equation gives
ln(x)=−ΔHfusR(1T2−1TM)ln(x)=−ΔHfusR(1T2−1TM)
where now TMTM is the melting temperature of the pure solute. As ΔH/TMΔH/TM is a constant then the mole fraction and hence solubility in solution is
ln(x)∝−ΔHfusR(1T2)ln(x)∝−ΔHfusR(1T2)
which shows that the mole fraction at temperature T varies as
xT∝exp(−ΔHfusRT)xT∝exp(−ΔHfusRT)
and so the solubility rises with increase in temperature. Different species will rise more or less slowly depending on their heat of fusion as ΔHfus/RΔHfus/R.
Using data for NaCl shows that the mole fraction hardly varies between 200200 to 400400 K, whereas there is a huge increase for NaNO3NaNOX3 under the same conditions.
Answered by
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NaCl is an ionic bond present is strong intramolecular force
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