3wat is
the percentage of equatorial form present in cyclohexanol at equilibrium
wat is
the value of equilibrium constant and Gibbs free energy change at equilibrium
wat is
the relation bet ke,kaandk
cyclo hexanol exists in equatorial and axial form in equilibrium with each other.
when acetylation of cyclohexanol is carried out and k is overall observed rate constant of acetylation and ke and ka are rate constant of acetylation of the equatorial and axial forms respectively the k=X1ke+X2ka. where X1 and X2 are mole fractions in equatorial and axial forms respectively. It is difficult to obtain valus of ke and ka but it has been determined by using cis and trans 4 t butyl cyclohexane derivatives, assuming that t butyl group is exclusively equatiorial k for cyclohexanol is 8.37*10-5,ka for cis 4 t butyl cyclohexanol with 100% axial form is 2.89*10-5 and similarly ke for trans 4 t butyl cyclohexanol is 10.65*10-5Lmol-1K-1
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1the answer are % equitorial is 29.38% relation after approx 100k=29Ke+71Ka and Keq.=8.4/100000 and ΔG=+23.4209592kJ/mole
3hey rahul how did u get the % equitorial is 29.38%
and the relation 100k=29Ke+71Ka
can others help me.
1hey sankara i can give u the soln for % case but for equilibria first tell me eq is for
equatorial →axial or
axial →equatorial
1k=X1ke+X2ka. and X1+X2=1
K=X1ke+Ka(1-X1)
K=8.37*10-5, Ke=10.65*10-5 Ka=2.89*10-5 putting the values u can get X1=.7061
so %equitorial form=70.6 %
1equilibriam cons for eq axial →equitorial Keq=%equtorial/%axial
on calculating Keq=2.4
relation 100k=29Ka+71Ke
ΔG=-2.303 RTlog Keq assuming T=300
ΔG=-2186.2 JULE/mole
1gagar has given sme sol. which was going to give hey gagar it is Joule yaar