PHYSICAL CHEM QUESTION: Consider cis and trans isomerization reaction (R=D) at 3

Question

PHYSICAL CHEM QUESTION:

Consider cis and trans isomerization reaction (R=D) at 300K :

C2H2D2 (cis) <<<>>> C2H2D2 (trans)

a) One mole of the cis gas is inroduced into a chamber and allowed to isomerize, forming the trans isomer. At equilibrium, there is an equimolar mixture the two forms. determine Delta G when the extent of the reaction is 0.1 (i.e. when the amount of trans formed is 0.1 moles)

b) What is Delta G for a mixture of 0.9 moles of trans and 0.1 moles of cis? what is the difference between this result and that of part a) and what does it mean?

Explanation / Answer

cis <=> trans

At equilibrium: moles of cis = moles of trans => [cis] = [trans]

Equilibrium constant K = [trans]/[cis] = 1

Temperature T = 300 K

Molar gas constant R = 8.314 J/mol.K


(a) When moles of trans 0.1 mol => moles of cis = 1 - 0.1 = 0.9 mol

Reaction quotient Q = [trans]/[cis] = moles of trans/moles of cis = 0.1/0.9 = 1/9


Delta G = Delta Go + RT ln Q = -RT ln K + RT ln Q

= -8.314 x 300 x ln(1) + 8.314 x 300 x ln(1/9)

= -5480 J = -5.48 kJ


(b) When moles of trans 0.9 mol and moles of cis = 0.1 mol

Reaction quotient Q = [trans]/[cis] = moles of trans/moles of cis = 0.9/0.1 = 9


Delta G = Delta Go + RT ln Q = -RT ln K + RT ln Q

= -8.314 x 300 x ln(1) + 8.314 x 300 x ln(9)

= +5480 J = +5.48 kJ


For part (a) => Delta G is negative => forward reaction is favored

=> reaction proceeds from cis => trans


For part (b) => Delta G is positive => reverse reaction is favored

=> reaction proceeds from trans => cis

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