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Please explain all options in image question.......).........)...#....

Sudhanva G V , 7 Years ago
Grade 11
anser 1 Answers
Samyak Jain
For reversible reactions, equilibrium contant K is related with temperature as
K = (Af /Ar) e–\DeltaH / RT , where Af = Arrhenius constant of forward reaction, Ar = Arrhenius constant for reverse reaction.
\DeltaH = (Ea)r – (Ea)f = Difference of activation energies of reverse and forward reactions.
For endothermic reaction, \DeltaH > 0.
lnK = ln(Af /Ar) + ln(e–\DeltaH / RT)  i.e.  lnK = ln(Af /Ar)  – \DeltaH / RT
lnK = (– \DeltaH / R)(1/T) + ln(Af /Ar)          ….(1).   Compare (1) with y = mx + c
We infer from above equation that lnK vs 1/T graph is a straight line with negative slope (–\DeltaH / R).
Differentiate (1) wrt x.
\Rightarrow d(lnK)/dT = (– \DeltaH / R)( –1/T2)       [\because (– \DeltaH / R) and ln(Af /Ar) are constants.]
d(lnK)/dT = \DeltaH / RT2 > 0  \because \DeltaH, R, T2 are positive.
If dT > 0, then d(lnK) >0 and if dT 0 i.e. if T is increased, K also increases.
This shows d(lnK) vs 1/T2 graph will have positive slope [equation is d(lnK) = (\DeltaH / R)(dT) (1/T2)].
Last Activity: 7 Years ago
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