Question
Question: Variation of K with temperature as given by Van’t Hoff equation can be written as: (A) \({ logK }_...
Variation of K with temperature as given by Van’t Hoff equation can be written as:
(A) logK2÷K1=−ΔH÷2.303[1÷T1−1÷T2]
(B) logK2÷K1=ΔH÷2.303[1÷T2−1÷T1]
(C) logK2÷K1=−ΔH÷2.303[1÷T2−1÷T1]
(D) None of the above
Solution
Hint: Van't Hoff factor i, determines the extent of association or dissociation, which can lead to the abnormal molar mass of a substance.
Complete step-by-step answer:
Van't Hoff equation: This equation gives the variation of the equilibrium constant of a reaction with temperature.
As we know that,
dlnKp÷dT=−ΔH−÷RT2
dlnKp=ΔH−÷R1/T2×dT
where, Kp = Equilibrium constant
R = Ideal gas constant
T = temperature
ΔH = standard enthalpy change
Integrating both the sides, we get
∫dlnKp=ΔH−÷R∫1÷T2.dT[lnKp]K1K2=ΔH−÷R[1/T]T1T2
lneK1K2ΔH−[1÷T1−1/T2]
Now, convert lne to log10 by multiplying it with 2.303,we get
2.303log10K1K2ΔH−÷R[T2−T1/T1T2]
Rearranging the values, we get
log10K1K2ΔH−÷R2.303[T2−T1/T1T2]
logK2÷K1=ΔH÷2.303[1÷T2−1÷T1]
Hence, the correct option is B.
Additional Information:
The van 't Hoff equation also known as the Vukancic-Vukovic equation in chemical thermodynamics relates the change in temperature T to the change in the equilibrium constant K, given the standard enthalpy change ΔH∘ for the process.
The van 't Hoff factor is a measure of the effect of a solute on colligative properties such as:
(I) Osmotic pressure,
(II) Relative lowering in vapor pressure,
(III) Elevation of boiling point and
(IV) Freezing point depression.
For substances which do not completely ionize in water for those substances, it is not an integer.
Compounds that completely ionize into solution, cations, and anions formed attract each other and form aggregates in concentrated solutions. ex- NaCl ionizes to Na+ and Cl− , at high concentrations of NaCl , its ions forms aggregates in solution, therefore i value is less than 2 in such case, 1<i<2,
To increase the value of i we take a dilute solution. For dilute solutions aggregations between cations and anions is less i=2.
In some cases, the substance added in the solvent forms self-aggregation, in that case also i < 1 as the number of moles dissolved in the solvent becomes less than the number of moles added due to self-association.
Note: The possibility to make a mistake is that you may choose option A. But there will be positive signs instead of negative signs as we when we integrate it will resolve.